Q2: Compute P(A B), if P(B) = 0.5 and P(A cap B) = 0.32

Given: P(B) = 0.5 P(A cap B) = 0.32 To Find: P(A B) Formula: Conditional Probability: P(A B) = P(A cap B)/P(B), provided P(B) ≠ 0. Solution: P(A B) = P(A cap B)/P(B) P(A B) = 0.32/0.5 = 32/50 = 16/25 = 0.64 Final Answer: P(A B) = 0.64

Q4: Evaluate P(A cup B), if 2P(A) = P(B) = 5/13 and P(A B) = 2/5

Given: 2P(A) = P(B) = 5/13 P(A B) = 2/5 To Find: P(A cup B) Solution: First, we find P(A) and P(B) from the given information. P(B) = 5/13 2P(A) = 5/13 Rightarrow P(A) = 5/26 Next, we use the conditional probability formula to find P(A cap B). P(A B) = P(A cap B)/P(B) P(A cap B) = P(A B) × P(B) P(A…

Q7 (i): E: tail appears on one coin, F: one coin shows head

Given: Two coins are tossed once. The sample space is: S = HH, HT, TH, TT Total number of outcomes, n(S) = 4. Each outcome has a probability of 1/4. (i) E: tail appears on one coin, F: one coin shows head 'Tail appears on one coin' means exactly one tail. Outcomes for event E: E = HT, TH. So, P(E)…

Q11 (i): P(E F) and P(F E)

Given: A fair die is rolled. The sample space is S = 1, 2, 3, 4, 5, 6. E = 1,3,5 Rightarrow P(E) = 3/6 = 1/2 F = 2,3 Rightarrow P(F) = 2/6 = 1/3 G = 2,3,4,5 Rightarrow P(G) = 4/6 = 2/3 (i) P(E F) and P(F E) E cap F = 3. So, P(E cap F) = 1/6. P(E F) = P(E cap F)/P(F) = 1/6/2/6 = 1/2 P(F E) = P(F cap…

Q11 (ii): P(E G) and P(G E)

Given: A fair die is rolled. The sample space is S = 1, 2, 3, 4, 5, 6. E = 1,3,5 Rightarrow P(E) = 3/6 = 1/2 F = 2,3 Rightarrow P(F) = 2/6 = 1/3 G = 2,3,4,5 Rightarrow P(G) = 4/6 = 2/3 (i) P(E F) and P(F E) E cap F = 3. So, P(E cap F) = 1/6. P(E F) = P(E cap F)/P(F) = 1/6/2/6 = 1/2 P(F E) = P(F cap…

Q11 (iii): P((E cup F) G) and P((E cap F) G)

Given: A fair die is rolled. The sample space is S = 1, 2, 3, 4, 5, 6. E = 1,3,5 Rightarrow P(E) = 3/6 = 1/2 F = 2,3 Rightarrow P(F) = 2/6 = 1/3 G = 2,3,4,5 Rightarrow P(G) = 4/6 = 2/3 (i) P(E F) and P(F E) E cap F = 3. So, P(E cap F) = 1/6. P(E F) = P(E cap F)/P(F) = 1/6/2/6 = 1/2 P(F E) = P(F cap…

Probability Class 12 Mathematics NCERT Solutions

Q1EXERCISE 13.1

Given that E and F are events such that $P(E) = 0.6$ , $P(F) = 0.3$ and $P(E \cap F) = 0.2$ , find $P(E|F)$ and $P(F|E)$

Solution

Given:

P(E) = 0.6

P(F) = 0.3

P(E \cap F) = 0.2

To Find:

P(E|F)

and

P(F|E)

Formula: Conditional Probability:

P(A|B) = \frac{P(A \cap B)}{P(B)}

, provided

P(B) \neq 0

.

Solution:

To find $P(E|F)$ : $P(E|F) = \frac{P(E \cap F)}{P(F)}$ $P(E|F) = \frac{0.2}{0.3} = \frac{2}{3}$
To find $P(F|E)$ : $P(F|E) = \frac{P(F \cap E)}{P(E)}$ Since $E \cap F = F \cap E$ , we have $P(E \cap F) = P(F \cap E)$ . $P(F|E) = \frac{0.2}{0.6} = \frac{2}{6} = \frac{1}{3}$

Final Answer:

P(E|F) = \frac{2}{3}

P(F|E) = \frac{1}{3}

Q2EXERCISE 13.1

Compute $P(A|B)$ , if $P(B) = 0.5$ and $P(A \cap B) = 0.32$

Solution

Given:

P(B) = 0.5

P(A \cap B) = 0.32

To Find:

P(A|B)

Formula: Conditional Probability:

P(A|B) = \frac{P(A \cap B)}{P(B)}

, provided

P(B) \neq 0

.

Solution:

P(A|B) = \frac{P(A \cap B)}{P(B)}

P(A|B) = \frac{0.32}{0.5} = \frac{32}{50} = \frac{16}{25} = 0.64

Final Answer:

P(A|B) = 0.64

Q3EXERCISE 13.1

If $P(A) = 0.8$ , $P(B) = 0.5$ and $P(B|A) = 0.4$ , find

(i)

$P(A \cap B)$

(ii)

$P(A|B)$

(iii)

$P(A \cup B)$

Solution

Given:

P(A) = 0.8

P(B) = 0.5

P(B|A) = 0.4

Solution:

(i) To find $P(A \cap B)$ We know that

P(B|A) = \frac{P(A \cap B)}{P(A)}

. Rearranging the formula gives the multiplication rule of probability:

P(A \cap B) = P(B|A) \times P(A)

P(A \cap B) = 0.4 \times 0.8 = 0.32

(ii) To find $P(A|B)$ Using the formula for conditional probability:

P(A|B) = \frac{P(A \cap B)}{P(B)}

P(A|B) = \frac{0.32}{0.5} = \frac{32}{50} = 0.64

(iii) To find $P(A \cup B)$ Using the addition rule of probability:

P(A \cup B) = P(A) + P(B) - P(A \cap B)

P(A \cup B) = 0.8 + 0.5 - 0.32

P(A \cup B) = 1.3 - 0.32 = 0.98

Final Answer:

(i)

P(A \cap B) = 0.32

(ii)

P(A|B) = 0.64

(iii)

P(A \cup B) = 0.98

Q4EXERCISE 13.1

Evaluate $P(A \cup B)$ , if $2P(A) = P(B) = \frac{5}{13}$ and $P(A|B) = \frac{2}{5}$

Solution

Given:

2P(A) = P(B) = \frac{5}{13}

P(A|B) = \frac{2}{5}

To Find:

P(A \cup B)

Solution: First, we find

P(A)

and

P(B)

from the given information.

P(B) = \frac{5}{13}

2P(A) = \frac{5}{13} \Rightarrow P(A) = \frac{5}{26}

Next, we use the conditional probability formula to find

P(A \cap B)

.

P(A|B) = \frac{P(A \cap B)}{P(B)}

P(A \cap B) = P(A|B) \times P(B)

P(A \cap B) = \frac{2}{5} \times \frac{5}{13} = \frac{2}{13}

Finally, we use the addition rule of probability to find

P(A \cup B)

.

P(A \cup B) = P(A) + P(B) - P(A \cap B)

P(A \cup B) = \frac{5}{26} + \frac{5}{13} - \frac{2}{13}

P(A \cup B) = \frac{5}{26} + \frac{10}{26} - \frac{4}{26}

P(A \cup B) = \frac{5 + 10 - 4}{26} = \frac{11}{26}

Final Answer:

P(A \cup B) = \frac{11}{26}

Q5EXERCISE 13.1

If $P(A) = \frac{6}{11}$ , $P(B) = \frac{5}{11}$ and $P(A \cup B) = \frac{7}{11}$ , find

(i)

$P(A \cap B)$

(ii)

$P(A|B)$

(iii)

$P(B|A)$

Solution

Given:

P(A) = \frac{6}{11}

P(B) = \frac{5}{11}

P(A \cup B) = \frac{7}{11}

Solution:

(i) To find $P(A \cap B)$ Using the addition rule of probability,

P(A \cup B) = P(A) + P(B) - P(A \cap B)

. Rearranging the formula:

P(A \cap B) = P(A) + P(B) - P(A \cup B)

P(A \cap B) = \frac{6}{11} + \frac{5}{11} - \frac{7}{11}

P(A \cap B) = \frac{11}{11} - \frac{7}{11} = \frac{4}{11}

(ii) To find $P(A|B)$ Using the conditional probability formula:

P(A|B) = \frac{P(A \cap B)}{P(B)}

P(A|B) = \frac{4/11}{5/11} = \frac{4}{5}

(iii) To find $P(B|A)$ Using the conditional probability formula:

P(B|A) = \frac{P(A \cap B)}{P(A)}

P(B|A) = \frac{4/11}{6/11} = \frac{4}{6} = \frac{2}{3}

Final Answer:

(i)

P(A \cap B) = \frac{4}{11}

(ii)

P(A|B) = \frac{4}{5}

(iii)

P(B|A) = \frac{2}{3}

Q6EXERCISE 13.1

A coin is tossed three times, where

(i)

E : head on third toss, F : heads on first two tosses

(ii)

E : at least two heads, F : at most two heads

(iii)

E : at most two tails, F : at least one tail Determine $P(E|F)$ in each case.

Solution

Given: A coin is tossed three times. The sample space is:

S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}

Total number of outcomes,

n(S) = 8

. Each outcome has a probability of

\frac{1}{8}

.

(i) E : head on third toss, F : heads on first two tosses Outcomes for event E:

E = \{HHH, HTH, TTH, THT\}

. So,

P(E) = \frac{4}{8} = \frac{1}{2}

. Outcomes for event F:

F = \{HHH, HHT\}

. So,

P(F) = \frac{2}{8} = \frac{1}{4}

. Outcomes for

E \cap F

:

E \cap F = \{HHH\}

. So,

P(E \cap F) = \frac{1}{8}

. Now, we find

P(E|F)

:

P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{1/8}{2/8} = \frac{1}{2}

(ii) E : at least two heads, F : at most two heads Outcomes for event E:

E = \{HHH, HHT, HTH, THH\}

. So,

P(E) = \frac{4}{8} = \frac{1}{2}

. Outcomes for event F:

F = \{HHT, HTH, THH, HTT, THT, TTH, TTT\}

. So,

P(F) = \frac{7}{8}

. Outcomes for

E \cap F

:

E \cap F = \{HHT, HTH, THH\}

. So,

P(E \cap F) = \frac{3}{8}

. Now, we find

P(E|F)

:

P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{3/8}{7/8} = \frac{3}{7}

(iii) E : at most two tails, F : at least one tail Outcomes for event E:

E = \{HHH, HHT, HTH, THH, HTT, THT, TTH\}

. So,

P(E) = \frac{7}{8}

. Outcomes for event F:

F = \{HHT, HTH, THH, HTT, THT, TTH, TTT\}

. So,

P(F) = \frac{7}{8}

. Outcomes for

E \cap F

:

E \cap F = \{HHT, HTH, THH, HTT, THT, TTH\}

. So,

P(E \cap F) = \frac{6}{8} = \frac{3}{4}

. Now, we find

P(E|F)

:

P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{6/8}{7/8} = \frac{6}{7}

Final Answer:

(i)

P(E|F) = \frac{1}{2}

(ii)

P(E|F) = \frac{3}{7}

(iii)

P(E|F) = \frac{6}{7}

Q7EXERCISE 13.1

Two coins are tossed once, where

(i)

E : tail appears on one coin, F : one coin shows head

(ii)

E : no tail appears, F : no head appears Determine $P(E|F)$ in each case.

Solution

Given: Two coins are tossed once. The sample space is:

S = \{HH, HT, TH, TT\}

Total number of outcomes,

n(S) = 4

. Each outcome has a probability of

\frac{1}{4}

.

(i) E : tail appears on one coin, F : one coin shows head 'Tail appears on one coin' means exactly one tail. Outcomes for event E:

E = \{HT, TH\}

. So,

P(E) = \frac{2}{4} = \frac{1}{2}

. 'One coin shows head' means exactly one head. Outcomes for event F:

F = \{HT, TH\}

. So,

P(F) = \frac{2}{4} = \frac{1}{2}

. Events E and F are the same. So,

E \cap F = \{HT, TH\}

. And

P(E \cap F) = \frac{2}{4} = \frac{1}{2}

. Now, we find

P(E|F)

:

P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{1/2}{1/2} = 1

(ii) E : no tail appears, F : no head appears Outcomes for event E:

E = \{HH\}

. So,

P(E) = \frac{1}{4}

. Outcomes for event F:

F = \{TT\}

. So,

P(F) = \frac{1}{4}

. Outcomes for

E \cap F

: There are no common outcomes.

E \cap F = \emptyset

. So,

P(E \cap F) = 0

. Now, we find

P(E|F)

:

P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{0}{1/4} = 0

Final Answer:

(i)

P(E|F) = 1

(ii)

P(E|F) = 0

Q8EXERCISE 13.1

A die is thrown three times, E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses Determine $P(E|F)$ .

Solution

Given: A die is thrown three times. The total number of outcomes in the sample space is

6 \times 6 \times 6 = 216

.

Event E: 4 appears on the third toss. The outcomes are of the form

(x, y, 4)

, where

x, y \in \{1, 2, 3, 4, 5, 6\}

. Number of outcomes for E,

n(E) = 6 \times 6 \times 1 = 36

.

Event F: 6 and 5 appears respectively on first two tosses. The outcomes are of the form

(6, 5, z)

, where

z \in \{1, 2, 3, 4, 5, 6\}

.

F = \{(6,5,1), (6,5,2), (6,5,3), (6,5,4), (6,5,5), (6,5,6)\}

. Number of outcomes for F,

n(F) = 6

. So,

P(F) = \frac{6}{216}

.

Event

E \cap F

: 4 on the third toss AND 6 and 5 on the first two tosses. There is only one such outcome:

(6, 5, 4)

. So,

E \cap F = \{(6, 5, 4)\}

. Number of outcomes for

E \cap F

,

n(E \cap F) = 1

. So,

P(E \cap F) = \frac{1}{216}

.

To Find:

P(E|F)

Solution:

P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{1/216}{6/216} = \frac{1}{6}

Alternatively, using number of outcomes:

P(E|F) = \frac{n(E \cap F)}{n(F)} = \frac{1}{6}

Final Answer:

P(E|F) = \frac{1}{6}

Q9EXERCISE 13.1

Mother, father and son line up at random for a family picture E : son on one end, F : father in middle Determine $P(E|F)$ .

Solution

Given: Mother (M), father (F), and son (S) line up at random. The sample space S consists of all possible arrangements (permutations).

S = \{MFS, MSF, FMS, FSM, SMF, SFM\}

Total number of outcomes,

n(S) = 3! = 6

.

Event E: son on one end. This means the son is either in the first or the last position.

E = \{SMF, SFM, MFS, FMS\}

. Number of outcomes for E,

n(E) = 4

.

Event F: father in middle. This means the arrangement is of the form (X, F, Y).

F = \{MFS, SFM\}

. Number of outcomes for F,

n(F) = 2

.

Event

E \cap F

: son on one end AND father in middle. We look for outcomes common to both E and F.

E \cap F = \{MFS, SFM\}

. Number of outcomes for

E \cap F

,

n(E \cap F) = 2

.

To Find:

P(E|F)

Solution: Since the event F has occurred, our reduced sample space is F itself, which has 2 outcomes: {MFS, SFM}. Out of these, the outcomes favorable to E are those where the son is on one end. Both {MFS} and {SFM} satisfy this condition. So, the number of favorable outcomes is 2.

P(E|F) = \frac{\text{Number of outcomes in } E \cap F}{\text{Number of outcomes in } F} = \frac{n(E \cap F)}{n(F)}

P(E|F) = \frac{2}{2} = 1

Final Answer:

P(E|F) = 1

Q10EXERCISE 13.1

A black and a red dice are rolled.

(a)

Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.

(b)

Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

Solution

Given: A black and a red die are rolled. The sample space S has

6 \times 6 = 36

outcomes. Let the outcome be represented by an ordered pair (black die, red die).

(a) Let E be the event 'obtaining a sum greater than 9'.

E = \{(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)\}

.

Let F be the event 'the black die resulted in a 5'.

F = \{(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}

.

n(F) = 6

.

We need to find

P(E|F)

. First, find

E \cap F

. These are outcomes in F where the sum is greater than 9.

E \cap F = \{(5,5), (5,6)\}

.

n(E \cap F) = 2

.

Now, we calculate the conditional probability:

P(E|F) = \frac{n(E \cap F)}{n(F)} = \frac{2}{6} = \frac{1}{3}

(b) Let A be the event 'obtaining the sum 8'.

A = \{(2,6), (3,5), (4,4), (5,3), (6,2)\}

.

Let B be the event 'the red die resulted in a number less than 4'. The red die can be 1, 2, or 3.

B = \{(1,1), (2,1), ..., (6,1), (1,2), (2,2), ..., (6,2), (1,3), (2,3), ..., (6,3)\}

.

n(B) = 6 \times 3 = 18

.

We need to find

P(A|B)

. First, find

A \cap B

. These are outcomes in A where the red die shows a number less than 4.

A \cap B = \{(5,3), (6,2)\}

.

n(A \cap B) = 2

.

Now, we calculate the conditional probability:

P(A|B) = \frac{n(A \cap B)}{n(B)} = \frac{2}{18} = \frac{1}{9}

Final Answer: (a)

\frac{1}{3}

(b)

\frac{1}{9}

Q11EXERCISE 13.1

A fair die is rolled. Consider events $E = \{1,3,5\}$ , $F = \{2,3\}$ and $G = \{2,3,4,5\}$ . Find

(i)

$P(E|F)$ and $P(F|E)$

(ii)

$P(E|G)$ and $P(G|E)$

(iii)

$P((E \cup F)|G)$ and $P((E \cap F)|G)$

Solution

Given: A fair die is rolled. The sample space is

S = \{1, 2, 3, 4, 5, 6\}

.

E = \{1,3,5\} \Rightarrow P(E) = \frac{3}{6} = \frac{1}{2}

F = \{2,3\} \Rightarrow P(F) = \frac{2}{6} = \frac{1}{3}

G = \{2,3,4,5\} \Rightarrow P(G) = \frac{4}{6} = \frac{2}{3}

(i) $P(E|F)$ and $P(F|E)$

E \cap F = \{3\}

. So,

P(E \cap F) = \frac{1}{6}

.

P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{1/6}{2/6} = \frac{1}{2}

P(F|E) = \frac{P(F \cap E)}{P(E)} = \frac{1/6}{3/6} = \frac{1}{3}

(ii) $P(E|G)$ and $P(G|E)$

E \cap G = \{3,5\}

. So,

P(E \cap G) = \frac{2}{6} = \frac{1}{3}

.

P(E|G) = \frac{P(E \cap G)}{P(G)} = \frac{2/6}{4/6} = \frac{2}{4} = \frac{1}{2}

P(G|E) = \frac{P(G \cap E)}{P(E)} = \frac{2/6}{3/6} = \frac{2}{3}

(iii) $P((E \cup F)|G)$ and $P((E \cap F)|G)$ First, find the required sets:

E \cup F = \{1, 2, 3, 5\}

.

(E \cup F) \cap G = \{2, 3, 5\}

. So,

P((E \cup F) \cap G) = \frac{3}{6} = \frac{1}{2}

.

E \cap F = \{3\}

.

(E \cap F) \cap G = \{3\}

. So,

P((E \cap F) \cap G) = \frac{1}{6}

.

Now, calculate the conditional probabilities:

P((E \cup F)|G) = \frac{P((E \cup F) \cap G)}{P(G)} = \frac{3/6}{4/6} = \frac{3}{4}

P((E \cap F)|G) = \frac{P((E \cap F) \cap G)}{P(G)} = \frac{1/6}{4/6} = \frac{1}{4}

Final Answer:

(i)

P(E|F) = \frac{1}{2}

,

P(F|E) = \frac{1}{3}

(ii)

P(E|G) = \frac{1}{2}

,

P(G|E) = \frac{2}{3}

(iii)

P((E \cup F)|G) = \frac{3}{4}

,

P((E \cap F)|G) = \frac{1}{4}

Q12EXERCISE 13.1

Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl, (ii) at least one is a girl?

Solution

Given: A family has two children. Let B denote a boy and G denote a girl. The sample space is:

S = \{BB, BG, GB, GG\}

, where the first letter represents the elder child. Total number of outcomes,

n(S) = 4

. Let E be the event that both children are girls.

E = \{GG\}

.

(i) Given that the youngest is a girl. Let F be the event that the youngest child is a girl.

F = \{BG, GG\}

.

n(F) = 2

. We need to find

P(E|F)

. First, find

E \cap F

. The outcome common to E and F is {GG}.

E \cap F = \{GG\}

.

n(E \cap F) = 1

.

The conditional probability is:

P(E|F) = \frac{n(E \cap F)}{n(F)} = \frac{1}{2}

(ii) Given that at least one is a girl. Let H be the event that at least one child is a girl.

H = \{BG, GB, GG\}

.

n(H) = 3

. We need to find

P(E|H)

. First, find

E \cap H

. The outcome common to E and H is {GG}.

E \cap H = \{GG\}

.

n(E \cap H) = 1

.

The conditional probability is:

P(E|H) = \frac{n(E \cap H)}{n(H)} = \frac{1}{3}

Final Answer:

(i)

The probability that both are girls given the youngest is a girl is

\frac{1}{2}

.

(ii)

The probability that both are girls given at least one is a girl is

\frac{1}{3}

.

Q13EXERCISE 13.1

An instructor has a question bank consisting of 300 easy True / False questions, 200 difficult True / False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?

Solution

Given: Total number of questions in the bank: Total = 300 (Easy T/F) + 200 (Difficult T/F) + 500 (Easy MCQ) + 400 (Difficult MCQ) = 1400 questions.

Let E be the event that the selected question is an 'easy' question. Let M be the event that the selected question is a 'multiple choice question' (MCQ).

We need to find the conditional probability

P(E|M)

.

Solution: First, let's find the number of outcomes for event M (the question is an MCQ). Number of MCQs = 500 (Easy) + 400 (Difficult) = 900. So,

n(M) = 900

.

Next, let's find the number of outcomes for the event

E \cap M

(the question is both easy and an MCQ). Number of easy MCQs = 500. So,

n(E \cap M) = 500

.

Now, we can calculate the conditional probability using the formula:

P(E|M) = \frac{n(E \cap M)}{n(M)}

P(E|M) = \frac{500}{900} = \frac{5}{9}

Final Answer: The probability that the selected question is easy given that it is a multiple choice question is

\frac{5}{9}

.

Q14EXERCISE 13.1

Given that the two numbers appearing on throwing two dice are different. Find the probability of the event 'the sum of numbers on the dice is 4'.

Solution

Given: Two dice are thrown. The total number of outcomes in the sample space is

6 \times 6 = 36

.

Let E be the event 'the sum of numbers on the dice is 4'.

E = \{(1,3), (2,2), (3,1)\}

.

Let F be the event 'the two numbers appearing are different'. The total number of outcomes is 36. The outcomes where the numbers are the same are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). There are 6 such outcomes. So, the number of outcomes where the numbers are different is

36 - 6 = 30

.

n(F) = 30

.

We need to find the conditional probability

P(E|F)

.

Solution: First, we find the number of outcomes for the event

E \cap F

(the sum is 4 AND the numbers are different). Looking at the outcomes for E, we exclude (2,2).

E \cap F = \{(1,3), (3,1)\}

.

n(E \cap F) = 2

.

Now, we calculate the conditional probability:

P(E|F) = \frac{n(E \cap F)}{n(F)}

P(E|F) = \frac{2}{30} = \frac{1}{15}

Final Answer: The probability that the sum is 4, given that the two numbers are different, is

\frac{1}{15}

.

Q15EXERCISE 13.1

Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event 'the coin shows a tail', given that 'at least one die shows a 3'.

Solution

Given: An experiment is performed as follows:

A die is thrown. Let the outcome be $d_1$ .
If $d_1 \in \{3, 6\}$ (a multiple of 3), the die is thrown again. Let the outcome be $d_2$ . The final outcome is $(d_1, d_2)$ .
If $d_1 \in \{1, 2, 4, 5\}$ , a coin is tossed. Let the outcome be $c$ . The final outcome is $(d_1, c)$ .

Let E be the event 'the coin shows a tail'. This can only happen if the first die roll is 1, 2, 4, or 5.

E = \{(1,T), (2,T), (4,T), (5,T)\}

.

Let F be the event 'at least one die shows a 3'. This can only happen if the first die roll is 3 or 6. If the first roll is 3, the outcomes are

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)

. If the first roll is 6, the second roll must be 3 for this event to occur, so the outcome is

(6,3)

.

F = \{(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (6,3)\}

.

We need to find the conditional probability

P(E|F)

.

Solution: To find

P(E|F)

, we need to find the intersection of events E and F, i.e.,

E \cap F

. Event E occurs only when a coin is tossed (first die roll is 1, 2, 4, or 5). Event F occurs only when a die is thrown a second time (first die roll is 3 or 6). These two sets of conditions are mutually exclusive. Therefore, the events E and F cannot happen at the same time.

E \cap F = \emptyset

(the empty set).

Since the intersection is empty, the probability of the intersection is zero.

P(E \cap F) = 0

.

The conditional probability is given by:

P(E|F) = \frac{P(E \cap F)}{P(F)}

Since

P(E \cap F) = 0

and

P(F) > 0

,

P(E|F) = \frac{0}{P(F)} = 0

Final Answer: The conditional probability of the event 'the coin shows a tail', given that 'at least one die shows a 3' is 0.

Q16EXERCISE 13.1

If $P(A) = \frac{1}{2}$ , $P(B) = 0$ , then $P(A|B)$ is

(A)

0

(B)

\frac{1}{2}

(C)

not defined

(D)

1

Solution

Given:

P(A) = \frac{1}{2}

P(B) = 0

To Find:

P(A|B)

Formula: The formula for conditional probability is:

P(A|B) = \frac{P(A \cap B)}{P(B)}

Explanation: The formula for conditional probability

P(A|B)

is defined only when the probability of the given event,

P(B)

, is not equal to zero. This is because division by zero is undefined. In this case,

P(B) = 0

. Therefore,

P(A|B)

is not defined.

Final Answer: The correct option is (C) not defined.

Q17EXERCISE 13.1

If A and B are events such that $P(A|B) = P(B|A)$ , then

(A)

A \subset B

but

A \neq B

(B)

A = B

(C)

A \cap B = \phi

(D)

P(A) = P(B)

Solution

Given: A and B are events such that

P(A|B) = P(B|A)

. We assume

P(A) \neq 0

and

P(B) \neq 0

for the conditional probabilities to be defined.

Solution: Using the definition of conditional probability:

P(A|B) = \frac{P(A \cap B)}{P(B)}

P(B|A) = \frac{P(B \cap A)}{P(A)}

Given that

P(A|B) = P(B|A)

, we can set the two expressions equal:

\frac{P(A \cap B)}{P(B)} = \frac{P(B \cap A)}{P(A)}

Since

A \cap B = B \cap A

, we have

P(A \cap B) = P(B \cap A)

. If

P(A \cap B) \neq 0

, we can simplify the equation:

\frac{1}{P(B)} = \frac{1}{P(A)}

This implies:

P(A) = P(B)

If

P(A \cap B) = 0

, then

P(A|B)=0

and

P(B|A)=0

, so the condition holds. In this case,

P(A)

and

P(B)

are not necessarily equal. However, among the given choices,

P(A) = P(B)

is the most general conclusion derived from the equality, especially when the events are not mutually exclusive.

Let's check the other options: (A) If

A \subset B

, then

P(B|A) = 1

and

P(A|B) = P(A)/P(B)

. For them to be equal,

P(A) = P(B)

, which would mean

A=B

(in terms of probability measure), contradicting

A \neq B

. (B) If

A=B

, then

P(A|A)=1

and

P(A|A)=1

. So

P(A|B)=P(B|A)

is true, and also

P(A)=P(B)

is true. This is a special case of (D). (C) If

A \cap B = \phi

, then

P(A \cap B) = 0

, so

P(A|B)=0

and

P(B|A)=0

. The condition holds, but does not require

P(A)=P(B)

.

Option (D) is the most direct algebraic consequence of the given statement under the common assumption that the events have a non-zero intersection.

Final Answer: The correct option is (D)

P(A) = P(B)

.

Q1EXERCISE 13.2

If $P(A) = \frac{3}{5}$ and $P(B) = \frac{1}{5}$ , find $P(A \cap B)$ if A and B are independent events.

Solution

Given:

P(A) = \frac{3}{5}

P(B) = \frac{1}{5}

A and B are independent events.

To Find:

P(A \cap B)

Formula: For independent events A and B, the probability of their intersection is given by:

P(A \cap B) = P(A) \times P(B)

Solution:

P(A \cap B) = \frac{3}{5} \times \frac{1}{5}

P(A \cap B) = \frac{3}{25}

Final Answer:

P(A \cap B) = \frac{3}{25}

Q2EXERCISE 13.2

Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.

Solution

Given: A pack of 52 playing cards. Two cards are drawn without replacement. Total cards = 52 Number of black cards = 26 Number of red cards = 26

To Find: The probability that both drawn cards are black.

Solution: Let E be the event that the first card drawn is black. Let F be the event that the second card drawn is black.

We need to find the probability of

E \cap F

, which is

P(E \cap F) = P(E) \times P(F|E)

.

Probability of the first card being black:

P(E) = \frac{\text{Number of black cards}}{\text{Total cards}} = \frac{26}{52} = \frac{1}{2}

After drawing one black card, there are 51 cards left in the pack, and 25 of them are black. So, the probability of the second card being black, given the first was black, is:

P(F|E) = \frac{\text{Remaining black cards}}{\text{Remaining total cards}} = \frac{25}{51}

Now, we use the multiplication rule:

P(E \cap F) = P(E) \times P(F|E) = \frac{1}{2} \times \frac{25}{51} = \frac{25}{102}

Final Answer: The probability that both cards are black is

\frac{25}{102}

.

Q3EXERCISE 13.2

A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.

Solution

Given: A box contains 15 oranges. Number of good oranges = 12 Number of bad oranges = 3 Three oranges are drawn without replacement. The box is approved if all three drawn oranges are good.

To Find: The probability that the box will be approved for sale.

Solution: Let

G_1

be the event that the first orange drawn is good. Let

G_2

be the event that the second orange drawn is good. Let

G_3

be the event that the third orange drawn is good.

The box is approved if the event

G_1 \cap G_2 \cap G_3

occurs. We use the multiplication rule for probability:

P(G_1 \cap G_2 \cap G_3) = P(G_1) \times P(G_2|G_1) \times P(G_3|G_1 \cap G_2)

Probability of the first orange being good:

P(G_1) = \frac{12}{15}

Given the first was good, there are 11 good oranges left out of 14 total oranges.

P(G_2|G_1) = \frac{11}{14}

Given the first two were good, there are 10 good oranges left out of 13 total oranges.

P(G_3|G_1 \cap G_2) = \frac{10}{13}

Now, we multiply these probabilities:

P(\text{approval}) = \frac{12}{15} \times \frac{11}{14} \times \frac{10}{13}

P(\text{approval}) = \frac{4}{5} \times \frac{11}{14} \times \frac{10}{13} = \frac{4}{5} \times \frac{11}{7} \times \frac{5}{13} = \frac{4 \times 11}{7 \times 13} = \frac{44}{91}

Final Answer: The probability that the box will be approved for sale is

\frac{44}{91}

.

Q4EXERCISE 13.2

A fair coin and an unbiased die are tossed. Let A be the event 'head appears on the coin' and B be the event '3 on the die'. Check whether A and B are independent events or not.

Solution

Given: A fair coin and an unbiased die are tossed. Event A: 'head appears on the coin'. Event B: '3 on the die'.

To Check: Whether A and B are independent events.

Condition for Independence: Two events A and B are independent if

P(A \cap B) = P(A) \times P(B)

.

Solution: The sample space for tossing a coin is {H, T}. The probability of getting a head is:

P(A) = \frac{1}{2}

The sample space for tossing a die is {1, 2, 3, 4, 5, 6}. The probability of getting a 3 is:

P(B) = \frac{1}{6}

Now, let's find the product of their probabilities:

P(A) \times P(B) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}

The event

A \cap B

is 'head appears on the coin AND 3 appears on the die'. The total sample space for the combined experiment has

2 \times 6 = 12

outcomes. The outcome corresponding to

A \cap B

is (H, 3). There is only 1 favorable outcome out of 12. So,

P(A \cap B) = \frac{1}{12}

.

Since

P(A \cap B) = P(A) \times P(B)

, the events A and B are independent.

Final Answer: Yes, A and B are independent events.

Q5EXERCISE 13.2

A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, 'the number is even,' and B be the event, 'the number is red'. Are A and B independent?

Solution

Given: A special die is tossed. The sample space is

S = \{1, 2, 3, 4, 5, 6\}

. Red numbers: {1, 2, 3} Green numbers: {4, 5, 6}

Event A: 'the number is even'.

A = \{2, 4, 6\}

. So,

P(A) = \frac{n(A)}{n(S)} = \frac{3}{6} = \frac{1}{2}

.

Event B: 'the number is red'.

B = \{1, 2, 3\}

. So,

P(B) = \frac{n(B)}{n(S)} = \frac{3}{6} = \frac{1}{2}

.

To Check: Are A and B independent?

Condition for Independence: Two events A and B are independent if

P(A \cap B) = P(A) \times P(B)

.

Solution: First, find the intersection

A \cap B

. This is the event 'the number is even AND red'.

A \cap B = \{2\}

. So,

P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{1}{6}

.

Next, find the product of the individual probabilities:

P(A) \times P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}

.

Now, compare

P(A \cap B)

and

P(A) \times P(B)

.

\frac{1}{6} \neq \frac{1}{4}

.

Since

P(A \cap B) \neq P(A) \times P(B)

, the events A and B are not independent.

Final Answer: No, A and B are not independent events.

Q6EXERCISE 13.2

Let E and F be events with $P(E) = \frac{3}{5}$ , $P(F) = \frac{3}{10}$ and $P(E \cap F) = \frac{1}{5}$ . Are E and F independent?

Solution

Given:

P(E) = \frac{3}{5}

P(F) = \frac{3}{10}

P(E \cap F) = \frac{1}{5}

To Check: Are E and F independent?

Condition for Independence: Two events E and F are independent if

P(E \cap F) = P(E) \times P(F)

.

Solution: Let's calculate the product of the individual probabilities:

P(E) \times P(F) = \frac{3}{5} \times \frac{3}{10} = \frac{9}{50}

Now, we compare this product with the given probability of the intersection,

P(E \cap F) = \frac{1}{5}

. To compare, let's express

\frac{1}{5}

with a denominator of 50:

\frac{1}{5} = \frac{1 \times 10}{5 \times 10} = \frac{10}{50}

.

We can see that:

\frac{9}{50} \neq \frac{10}{50}

Therefore,

P(E) \times P(F) \neq P(E \cap F)

.

Since the condition for independence is not met, the events E and F are not independent.

Final Answer: No, E and F are not independent.

Q7EXERCISE 13.2

Given that the events A and B are such that $P(A) = \frac{1}{2}$ , $P(A \cup B) = \frac{3}{5}$ and $P(B) = p$ . Find $p$ if they are (i) mutually exclusive (ii) independent.

Solution

Given:

P(A) = \frac{1}{2}

P(A \cup B) = \frac{3}{5}

P(B) = p

(i) A and B are mutually exclusive If events A and B are mutually exclusive, it means they cannot occur at the same time, so their intersection is empty:

A \cap B = \emptyset

. This implies

P(A \cap B) = 0

.

Using the addition rule of probability:

P(A \cup B) = P(A) + P(B) - P(A \cap B)

\frac{3}{5} = \frac{1}{2} + p - 0

p = \frac{3}{5} - \frac{1}{2}

p = \frac{6 - 5}{10} = \frac{1}{10}

(ii) A and B are independent If events A and B are independent, then

P(A \cap B) = P(A) \times P(B)

.

P(A \cap B) = \frac{1}{2} \times p = \frac{p}{2}

.

Using the addition rule of probability:

P(A \cup B) = P(A) + P(B) - P(A \cap B)

\frac{3}{5} = \frac{1}{2} + p - \frac{p}{2}

\frac{3}{5} = \frac{1}{2} + \frac{p}{2}

\frac{p}{2} = \frac{3}{5} - \frac{1}{2}

\frac{p}{2} = \frac{6 - 5}{10} = \frac{1}{10}

p = 2 \times \frac{1}{10} = \frac{2}{10} = \frac{1}{5}

Final Answer:

(i)

If the events are mutually exclusive,

p = \frac{1}{10}

.

(ii)

If the events are independent,

p = \frac{1}{5}

.

Q8EXERCISE 13.2

Let A and B be independent events with $P(A) = 0.3$ and $P(B) = 0.4$ . Find

(i)

$P(A \cap B)$

(ii)

$P(A \cup B)$

(iii)

$P(A|B)$

(iv)

$P(B|A)$

Solution

Given: A and B are independent events.

P(A) = 0.3

P(B) = 0.4

Solution:

(i) $P(A \cap B)$ For independent events,

P(A \cap B) = P(A) \times P(B)

.

P(A \cap B) = 0.3 \times 0.4 = 0.12

(ii) $P(A \cup B)$ Using the addition rule,

P(A \cup B) = P(A) + P(B) - P(A \cap B)

.

P(A \cup B) = 0.3 + 0.4 - 0.12

P(A \cup B) = 0.7 - 0.12 = 0.58

(iii) $P(A|B)$ For independent events, the occurrence of B does not affect the probability of A. Therefore,

P(A|B) = P(A)

.

P(A|B) = 0.3

Alternatively, using the formula:

P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.12}{0.4} = \frac{12}{40} = \frac{3}{10} = 0.3

.

(iv) $P(B|A)$ For independent events, the occurrence of A does not affect the probability of B. Therefore,

P(B|A) = P(B)

.

P(B|A) = 0.4

Alternatively, using the formula:

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.12}{0.3} = \frac{12}{30} = \frac{4}{10} = 0.4

.

Final Answer:

(i)

P(A \cap B) = 0.12

(ii)

P(A \cup B) = 0.58

(iii)

P(A|B) = 0.3

(iv)

P(B|A) = 0.4

Q9EXERCISE 13.2

If A and B are two events such that $P(A) = \frac{1}{4}$ , $P(B) = \frac{1}{2}$ and $P(A \cap B) = \frac{1}{8}$ , find P(not A and not B).

Solution

Given:

P(A) = \frac{1}{4}

P(B) = \frac{1}{2}

P(A \cap B) = \frac{1}{8}

To Find: P(not A and not B), which is denoted as

P(A' \cap B')

.

Solution: First, let's check if events A and B are independent.

P(A) \times P(B) = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}

Since

P(A) \times P(B) = P(A \cap B)

, the events A and B are independent.

If A and B are independent events, then their complements A' and B' are also independent. Therefore,

P(A' \cap B') = P(A') \times P(B')

.

First, we find the probabilities of the complements:

P(A') = 1 - P(A) = 1 - \frac{1}{4} = \frac{3}{4}

P(B') = 1 - P(B) = 1 - \frac{1}{2} = \frac{1}{2}

Now, we can find

P(A' \cap B')

:

P(A' \cap B') = P(A') \times P(B') = \frac{3}{4} \times \frac{1}{2} = \frac{3}{8}

Alternative Method (using De Morgan's Law): P(not A and not B) =

P(A' \cap B')

. By De Morgan's Law,

A' \cap B' = (A \cup B)'

. So,

P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)

. First, we find

P(A \cup B)

:

P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8} = \frac{2}{8} + \frac{4}{8} - \frac{1}{8} = \frac{5}{8}

. Then,

P(A' \cap B') = 1 - \frac{5}{8} = \frac{3}{8}

.

Final Answer: P(not A and not B) =

\frac{3}{8}

.

Q10EXERCISE 13.2

Events A and B are such that $P(A) = \frac{1}{2}$ , $P(B) = \frac{7}{12}$ and P(not A or not B) = $\frac{1}{4}$ . State whether A and B are independent ?

Solution

Given:

P(A) = \frac{1}{2}

P(B) = \frac{7}{12}

P(not A or not B) =

P(A' \cup B') = \frac{1}{4}

To Check: Are A and B independent?

Condition for Independence: Two events A and B are independent if

P(A \cap B) = P(A) \times P(B)

.

Solution: First, we need to find

P(A \cap B)

from the given information. We are given

P(A' \cup B')

. By De Morgan's Law,

A' \cup B' = (A \cap B)'

. So,

P(A' \cup B') = P((A \cap B)') = \frac{1}{4}

.

The probability of an event and its complement sum to 1:

P(A \cap B) + P((A \cap B)') = 1

P(A \cap B) + \frac{1}{4} = 1

P(A \cap B) = 1 - \frac{1}{4} = \frac{3}{4}

.

Now, let's calculate the product of the individual probabilities:

P(A) \times P(B) = \frac{1}{2} \times \frac{7}{12} = \frac{7}{24}

Finally, we compare

P(A \cap B)

with

P(A) \times P(B)

. Is

\frac{3}{4} = \frac{7}{24}

? To compare, we can use a common denominator of 24:

\frac{3}{4} = \frac{3 \times 6}{4 \times 6} = \frac{18}{24}

.

Clearly,

\frac{18}{24} \neq \frac{7}{24}

. Since

P(A \cap B) \neq P(A) \times P(B)

, the events A and B are not independent.

Final Answer: A and B are not independent events.

Q11EXERCISE 13.2

Given two independent events A and B such that $P(A) = 0.3$ , $P(B) = 0.6$ . Find

(i)

$P(A \text{ and } B)$

(ii)

$P(A \text{ and not } B)$

(iii)

$P(A \text{ or } B)$

(iv)

P(neither A nor B)

Solution

Given: A and B are independent events.

P(A) = 0.3

P(B) = 0.6

Solution:

(i) $P(A \text{ and } B)$ This is

P(A \cap B)

. Since A and B are independent:

P(A \cap B) = P(A) \times P(B) = 0.3 \times 0.6 = 0.18

(ii) $P(A \text{ and not } B)$ This is

P(A \cap B')

. If A and B are independent, then A and B' are also independent. First, find

P(B') = 1 - P(B) = 1 - 0.6 = 0.4

.

P(A \cap B') = P(A) \times P(B') = 0.3 \times 0.4 = 0.12

(iii) $P(A \text{ or } B)$ This is

P(A \cup B)

. Using the addition rule:

P(A \cup B) = P(A) + P(B) - P(A \cap B)

P(A \cup B) = 0.3 + 0.6 - 0.18 = 0.9 - 0.18 = 0.72

(iv) P(neither A nor B) This is

P(A' \text{ and } B')

, or

P(A' \cap B')

. If A and B are independent, then A' and B' are also independent. First, find

P(A') = 1 - P(A) = 1 - 0.3 = 0.7

. We already have

P(B') = 0.4

.

P(A' \cap B') = P(A') \times P(B') = 0.7 \times 0.4 = 0.28

Alternatively, using De Morgan's Law:

P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B) = 1 - 0.72 = 0.28

.

Final Answer:

(i)

P(A \text{ and } B) = 0.18

(ii)

P(A \text{ and not } B) = 0.12

(iii)

P(A \text{ or } B) = 0.72

(iv)

P(neither A nor B) =

0.28

Q12EXERCISE 13.2

A die is tossed thrice. Find the probability of getting an odd number at least once.

Solution

Given: A die is tossed three times.

To Find: The probability of getting an odd number at least once.

Solution: The outcomes of the three tosses are independent events. For a single toss of a fair die, the sample space is

S = \{1, 2, 3, 4, 5, 6\}

. Odd numbers: {1, 3, 5} Even numbers: {2, 4, 6}

Probability of getting an odd number in one toss:

P(\text{odd}) = \frac{3}{6} = \frac{1}{2}

. Probability of getting an even number in one toss:

P(\text{even}) = \frac{3}{6} = \frac{1}{2}

.

The event 'getting an odd number at least once' is the complement of the event 'getting no odd numbers at all'. 'Getting no odd numbers' means getting an even number on all three tosses.

Let A be the event 'getting an odd number at least once'. Let A' be the event 'getting an even number on all three tosses'. Then,

P(A) = 1 - P(A')

.

Since the tosses are independent, the probability of getting an even number on all three tosses is:

P(A') = P(\text{even on 1st}) \times P(\text{even on 2nd}) \times P(\text{even on 3rd})

P(A') = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}

Now, we can find the required probability:

P(A) = 1 - P(A') = 1 - \frac{1}{8} = \frac{7}{8}

Final Answer: The probability of getting an odd number at least once is

\frac{7}{8}

.

Q13EXERCISE 13.2

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that

(i)

both balls are red.

(ii)

first ball is black and second is red.

(iii)

one of them is black and other is red.

Solution

Given: A box contains 10 black (B) and 8 red (R) balls. Total balls =

10 + 8 = 18

. Two balls are drawn with replacement, which means the draws are independent events.

Probability of drawing a black ball:

P(B) = \frac{10}{18} = \frac{5}{9}

. Probability of drawing a red ball:

P(R) = \frac{8}{18} = \frac{4}{9}

.

Solution:

(i) both balls are red This means the first is red AND the second is red. Since the draws are independent:

P(R_1 \cap R_2) = P(R) \times P(R) = \frac{4}{9} \times \frac{4}{9} = \frac{16}{81}

(ii) first ball is black and second is red This is the event

B_1 \cap R_2

. Since the draws are independent:

P(B_1 \cap R_2) = P(B) \times P(R) = \frac{5}{9} \times \frac{4}{9} = \frac{20}{81}

(iii) one of them is black and other is red This can happen in two mutually exclusive ways:

First is black and second is red ( $B_1 \cap R_2$ ).
First is red and second is black ( $R_1 \cap B_2$ ).

The probability of this event is the sum of the probabilities of these two cases.

P(\text{one B and one R}) = P(B_1 \cap R_2) + P(R_1 \cap B_2)

We already calculated

P(B_1 \cap R_2) = \frac{20}{81}

. Now, for the second case:

P(R_1 \cap B_2) = P(R) \times P(B) = \frac{4}{9} \times \frac{5}{9} = \frac{20}{81}

.

Adding them up:

P(\text{one B and one R}) = \frac{20}{81} + \frac{20}{81} = \frac{40}{81}

Final Answer:

(i)

\frac{16}{81}

(ii)

\frac{20}{81}

(iii)

\frac{40}{81}

Q14EXERCISE 13.2

Probability of solving specific problem independently by A and B are $\frac{1}{2}$ and $\frac{1}{3}$ respectively. If both try to solve the problem independently, find the probability that

(i)

the problem is solved

(ii)

exactly one of them solves the problem.

Solution

Given: Let A be the event that person A solves the problem. Let B be the event that person B solves the problem.

P(A) = \frac{1}{2}

P(B) = \frac{1}{3}

The events are independent.

We can also find the probabilities of them not solving the problem:

P(A') = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2}

.

P(B') = 1 - P(B) = 1 - \frac{1}{3} = \frac{2}{3}

.

Solution:

(i) the problem is solved The problem is solved if at least one of them solves it. This is the event

A \cup B

. It is easier to calculate this using the complement: 1 - (probability that neither solves the problem). The event 'neither solves the problem' is

A' \cap B'

. Since A and B are independent, A' and B' are also independent.

P(A' \cap B') = P(A') \times P(B') = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}

Therefore, the probability that the problem is solved is:

P(A \cup B) = 1 - P(A' \cap B') = 1 - \frac{1}{3} = \frac{2}{3}

(ii) exactly one of them solves the problem This can happen in two mutually exclusive ways:

A solves the problem and B does not ( $A \cap B'$ ).
B solves the problem and A does not ( $A' \cap B$ ).

The total probability is the sum of the probabilities of these two cases.

P(\text{exactly one solves}) = P(A \cap B') + P(A' \cap B)

Since the events are independent:

P(A \cap B') = P(A) \times P(B') = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}

P(A' \cap B) = P(A') \times P(B) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}

Adding them up:

P(\text{exactly one solves}) = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}

Final Answer:

(i)

The probability that the problem is solved is

\frac{2}{3}

.

(ii)

The probability that exactly one of them solves the problem is

\frac{1}{2}

.

Q15EXERCISE 13.2

One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent ?

(i)

E : 'the card drawn is a spade' F : 'the card drawn is an ace'

(ii)

E : 'the card drawn is black' F : 'the card drawn is a king'

(iii)

E : 'the card drawn is a king or queen' F : 'the card drawn is a queen or jack'.

Solution

Condition for Independence: Two events E and F are independent if

P(E \cap F) = P(E) \times P(F)

. Total cards = 52.

(i) E: 'spade', F: 'ace'

P(E) = P(\text{spade}) = \frac{13}{52} = \frac{1}{4}

.

P(F) = P(\text{ace}) = \frac{4}{52} = \frac{1}{13}

.

E \cap F

: 'the card is the ace of spades'. There is only 1 such card.

P(E \cap F) = \frac{1}{52}

. Check the condition:

P(E) \times P(F) = \frac{1}{4} \times \frac{1}{13} = \frac{1}{52}

. Since

P(E \cap F) = P(E) \times P(F)

, the events are independent.

(ii) E: 'black', F: 'king'

P(E) = P(\text{black}) = \frac{26}{52} = \frac{1}{2}

.

P(F) = P(\text{king}) = \frac{4}{52} = \frac{1}{13}

.

E \cap F

: 'the card is a black king' (king of spades or king of clubs). There are 2 such cards.

P(E \cap F) = \frac{2}{52} = \frac{1}{26}

. Check the condition:

P(E) \times P(F) = \frac{1}{2} \times \frac{1}{13} = \frac{1}{26}

. Since

P(E \cap F) = P(E) \times P(F)

, the events are independent.

(iii) E: 'king or queen', F: 'queen or jack'

P(E) = P(\text{king or queen}) = \frac{4+4}{52} = \frac{8}{52} = \frac{2}{13}

.

P(F) = P(\text{queen or jack}) = \frac{4+4}{52} = \frac{8}{52} = \frac{2}{13}

.

E \cap F

: 'the card is a queen'. There are 4 queens.

P(E \cap F) = \frac{4}{52} = \frac{1}{13}

. Check the condition:

P(E) \times P(F) = \frac{2}{13} \times \frac{2}{13} = \frac{4}{169}

. Since

\frac{1}{13} \neq \frac{4}{169}

, the events are not independent.

Final Answer:

(i)

Independent

(ii)

Independent

(iii)

Not independent

Q16EXERCISE 13.2

In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random.

(a)

Find the probability that she reads neither Hindi nor English newspapers.

(b)

If she reads Hindi newspaper, find the probability that she reads English newspaper.

(c)

If she reads English newspaper, find the probability that she reads Hindi newspaper.

Solution

Given: Let H be the event that a student reads Hindi newspaper. Let E be the event that a student reads English newspaper.

P(H) = 60\% = 0.6

P(E) = 40\% = 0.4

P(H \cap E) = 20\% = 0.2

Solution:

(a) Find the probability that she reads neither Hindi nor English newspapers. This is the probability of the event

H' \cap E'

, which is the complement of the event

H \cup E

.

P(H' \cap E') = P((H \cup E)') = 1 - P(H \cup E)

. First, we find

P(H \cup E)

using the addition rule:

P(H \cup E) = P(H) + P(E) - P(H \cap E)

P(H \cup E) = 0.6 + 0.4 - 0.2 = 1.0 - 0.2 = 0.8

Now, we find the required probability:

P(H' \cap E') = 1 - 0.8 = 0.2

(b) If she reads Hindi newspaper, find the probability that she reads English newspaper. This is the conditional probability

P(E|H)

.

P(E|H) = \frac{P(E \cap H)}{P(H)}

P(E|H) = \frac{0.2}{0.6} = \frac{2}{6} = \frac{1}{3}

(c) If she reads English newspaper, find the probability that she reads Hindi newspaper. This is the conditional probability

P(H|E)

.

P(H|E) = \frac{P(H \cap E)}{P(E)}

P(H|E) = \frac{0.2}{0.4} = \frac{2}{4} = \frac{1}{2}

Final Answer: (a) The probability that she reads neither newspaper is 0.2 or 20%. (b) The probability that she reads English given she reads Hindi is

\frac{1}{3}

. (c) The probability that she reads Hindi given she reads English is

\frac{1}{2}

.

Q17EXERCISE 13.2

The probability of obtaining an even prime number on each die, when a pair of dice is rolled is

(A)

0

(B)

\frac{1}{3}

(C)

\frac{1}{12}

(D)

\frac{1}{36}

Solution

Given: A pair of dice is rolled.

To Find: The probability of obtaining an even prime number on each die.

Solution: The only even prime number is 2. So, the event is equivalent to 'getting a 2 on the first die AND getting a 2 on the second die'.

Let A be the event 'getting a 2 on the first die'. The probability of getting a 2 on a single die roll is

\frac{1}{6}

. So,

P(A) = \frac{1}{6}

.

Let B be the event 'getting a 2 on the second die'. Similarly,

P(B) = \frac{1}{6}

.

The two rolls are independent events. Therefore, the probability of both events occurring is the product of their individual probabilities.

P(A \cap B) = P(A) \times P(B) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}

Final Answer: The correct option is (D)

\frac{1}{36}

.

Q18EXERCISE 13.2

Two events A and B will be independent, if

(A)

A and B are mutually exclusive

(B)

P(A'B') = [1-P(A)][1-P(B)]

(C)

P(A) = P(B)

(D)

P(A) + P(B) = 1

Solution

To determine the condition for independence of two events A and B.

Definition of Independence: Two events A and B are independent if and only if

P(A \cap B) = P(A) \times P(B)

.

Let's analyze the given options:

(A) A and B are mutually exclusive If A and B are mutually exclusive (and have non-zero probabilities), then

P(A \cap B) = 0

. For them to be independent, we would need

P(A) \times P(B) = 0

, which implies either

P(A)=0

or

P(B)=0

. So, mutually exclusive events with non-zero probabilities cannot be independent. This statement is incorrect.

(B) $P(A'B') = [1-P(A)][1-P(B)]$ Here,

A'B'

denotes the intersection

A' \cap B'

. The expression can be written as

P(A' \cap B') = P(A') \times P(B')

, since

P(A') = 1 - P(A)

and

P(B') = 1 - P(B)

. A key property of independent events is that if A and B are independent, then their complements A' and B' are also independent. Conversely, if A' and B' are independent, then A and B are also independent. Therefore, the condition

P(A' \cap B') = P(A')P(B')

is equivalent to the condition that A and B are independent. This statement is correct.

(C) $P(A) = P(B)$ This condition is not related to independence. Two events can have equal probabilities but be dependent, or have unequal probabilities and be independent.

(D) $P(A) + P(B) = 1$ This condition is not related to independence. For example, if A and B are complements (

B=A'

), then

P(A)+P(B)=1

, but they are not independent (unless

P(A)

is 0 or 1).

Conclusion: The condition given in option (B) is a direct and equivalent statement for the independence of events A and B.

Final Answer: The correct option is (B)

P(A'B') = [1-P(A)][1-P(B)]

.

Q1EXERCISE 13.3

An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?

Solution

Given: Urn contains 5 Red (R) and 5 Black (B) balls. Total = 10 balls.

Let:

E_1

: The first ball drawn is red.

E_2

: The first ball drawn is black. A: The second ball drawn is red.

To Find:

P(A)

Solution: We use the Theorem of Total Probability:

P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2)

.

Step 1: Find probabilities of the first draw.

P(E_1) = P(\text{first is red}) = \frac{5}{10} = \frac{1}{2}

.

P(E_2) = P(\text{first is black}) = \frac{5}{10} = \frac{1}{2}

.

Step 2: Find conditional probabilities of the second draw. If the first ball drawn was red (

E_1

occurred), it is returned and 2 more red balls are added. The urn now contains:

5+2=7

Red balls

5

Black balls Total = 12 balls. The probability of drawing a red ball now is

P(A|E_1) = \frac{7}{12}

.

If the first ball drawn was black (

E_2

occurred), it is returned and 2 more black balls are added. The urn now contains:

5

Red balls

5+2=7

Black balls Total = 12 balls. The probability of drawing a red ball now is

P(A|E_2) = \frac{5}{12}

.

Step 3: Apply the Theorem of Total Probability.

P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2)

P(A) = \left(\frac{1}{2} \times \frac{7}{12}\right) + \left(\frac{1}{2} \times \frac{5}{12}\right)

P(A) = \frac{7}{24} + \frac{5}{24} = \frac{12}{24} = \frac{1}{2}

Final Answer: The probability that the second ball is red is

\frac{1}{2}

.

Q2EXERCISE 13.3

A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Solution

Given: Bag I: 4 Red (R), 4 Black (B) balls. Total = 8. Bag II: 2 Red (R), 6 Black (B) balls. Total = 8.

Let:

E_1

: Bag I is selected.

E_2

: Bag II is selected. A: The ball drawn is red.

To Find:

P(E_1|A)

, the probability that the ball was drawn from Bag I given that it is red.

Solution: We use Bayes' Theorem:

P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}

.

Step 1: Probabilities of selecting the bags. Since one bag is selected at random:

P(E_1) = \frac{1}{2}

P(E_2) = \frac{1}{2}

Step 2: Conditional probabilities of drawing a red ball. If Bag I is selected, the probability of drawing a red ball is:

P(A|E_1) = \frac{4}{8} = \frac{1}{2}

. If Bag II is selected, the probability of drawing a red ball is:

P(A|E_2) = \frac{2}{8} = \frac{1}{4}

.

Step 3: Apply Bayes' Theorem.

P(E_1|A) = \frac{\frac{1}{2} \times \frac{1}{2}}{\left(\frac{1}{2} \times \frac{1}{2}\right) + \left(\frac{1}{2} \times \frac{1}{4}\right)}

P(E_1|A) = \frac{\frac{1}{4}}{\frac{1}{4} + \frac{1}{8}} = \frac{\frac{1}{4}}{\frac{2+1}{8}} = \frac{\frac{1}{4}}{\frac{3}{8}}

P(E_1|A) = \frac{1}{4} \times \frac{8}{3} = \frac{2}{3}

Final Answer: The probability that the ball is drawn from the first bag is

\frac{2}{3}

.

Q3EXERCISE 13.3

Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostlier?

Solution

Given: Let H be the event that a student is a hostlier. Let D be the event that a student is a day scholar. Let A be the event that a student attains an A grade.

P(H) = 60\% = 0.6

P(D) = 40\% = 0.4

P(A|H) = 30\% = 0.3

(Probability of A grade given the student is a hostlier)

P(A|D) = 20\% = 0.2

(Probability of A grade given the student is a day scholar)

To Find:

P(H|A)

, the probability that a student is a hostlier given that he has an A grade.

Solution: We use Bayes' Theorem:

P(H|A) = \frac{P(H)P(A|H)}{P(H)P(A|H) + P(D)P(A|D)}

.

The denominator is the total probability of getting an A grade,

P(A)

.

Step 1: Calculate the numerator.

P(H)P(A|H) = 0.6 \times 0.3 = 0.18

.

Step 2: Calculate the denominator.

P(A) = P(H)P(A|H) + P(D)P(A|D)

P(A) = (0.6 \times 0.3) + (0.4 \times 0.2)

P(A) = 0.18 + 0.08 = 0.26

.

Step 3: Apply Bayes' Theorem.

P(H|A) = \frac{0.18}{0.26} = \frac{18}{26} = \frac{9}{13}

Final Answer: The probability that the student is a hostlier, given that he has an A grade, is

\frac{9}{13}

.

Q4EXERCISE 13.3

In answering a question on a multiple choice test, a student either knows the answer or guesses. Let $\frac{3}{4}$ be the probability that he knows the answer and $\frac{1}{4}$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $\frac{1}{4}$ . What is the probability that the student knows the answer given that he answered it correctly?

Solution

Given: Let K be the event that the student knows the answer. Let G be the event that the student guesses the answer. Let C be the event that the student answers correctly.

P(K) = \frac{3}{4}

P(G) = \frac{1}{4}

If the student knows the answer, he will answer it correctly. So, the probability of answering correctly given that he knows the answer is 1.

P(C|K) = 1

.

The probability of answering correctly given that he guesses is given as

\frac{1}{4}

.

P(C|G) = \frac{1}{4}

.

To Find:

P(K|C)

, the probability that the student knows the answer given that he answered it correctly.

Solution: We use Bayes' Theorem:

P(K|C) = \frac{P(K)P(C|K)}{P(K)P(C|K) + P(G)P(C|G)}

.

Step 1: Calculate the numerator.

P(K)P(C|K) = \frac{3}{4} \times 1 = \frac{3}{4}

.

Step 2: Calculate the denominator (total probability of answering correctly).

P(C) = P(K)P(C|K) + P(G)P(C|G)

P(C) = \left(\frac{3}{4} \times 1\right) + \left(\frac{1}{4} \times \frac{1}{4}\right)

P(C) = \frac{3}{4} + \frac{1}{16} = \frac{12}{16} + \frac{1}{16} = \frac{13}{16}

.

Step 3: Apply Bayes' Theorem.

P(K|C) = \frac{3/4}{13/16} = \frac{3}{4} \times \frac{16}{13} = \frac{3 \times 4}{13} = \frac{12}{13}

Final Answer: The probability that the student knows the answer given that he answered it correctly is

\frac{12}{13}

.

Q5EXERCISE 13.3

A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e. if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive ?

Solution

Given: Let D be the event that a person has the disease. Let H be the event that a person is healthy (

D'

). Let P be the event that the test result is positive.

From the problem statement:

P(D) = 0.1\% = \frac{0.1}{100} = 0.001

.

P(H) = 1 - P(D) = 1 - 0.001 = 0.999

.

The test is 99% effective in detecting the disease when present:

P(P|D) = 99\% = 0.99

. False positive rate for a healthy person:

P(P|H) = 0.5\% = \frac{0.5}{100} = 0.005

.

To Find:

P(D|P)

, the probability that a person has the disease given that his test result is positive.

Solution: We use Bayes' Theorem:

P(D|P) = \frac{P(D)P(P|D)}{P(D)P(P|D) + P(H)P(P|H)}

.

Step 1: Calculate the numerator.

P(D)P(P|D) = 0.001 \times 0.99 = 0.00099

.

Step 2: Calculate the denominator (total probability of a positive test).

P(P) = P(D)P(P|D) + P(H)P(P|H)

P(P) = (0.001 \times 0.99) + (0.999 \times 0.005)

P(P) = 0.00099 + 0.004995 = 0.005985

.

Step 3: Apply Bayes' Theorem.

P(D|P) = \frac{0.00099}{0.005985} = \frac{990}{5985}

To simplify the fraction:

P(D|P) = \frac{990 \div 45}{5985 \div 45} = \frac{22}{133}

Final Answer: The probability that a person has the disease given that his test result is positive is

\frac{22}{133}

(approximately 0.165).

Q6EXERCISE 13.3

There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin ?

Solution

Given: Three coins with different properties.

Let:

E_1

: The chosen coin is the two-headed coin.

E_2

: The chosen coin is the biased coin.

E_3

: The chosen coin is the unbiased coin. A: The toss results in a head.

Probabilities of choosing each coin: Since a coin is chosen at random, the probabilities are equal.

P(E_1) = \frac{1}{3}

P(E_2) = \frac{1}{3}

P(E_3) = \frac{1}{3}

Conditional probabilities of getting a head: If the two-headed coin is chosen:

P(A|E_1) = 1

. If the biased coin is chosen:

P(A|E_2) = 75\% = 0.75 = \frac{3}{4}

. If the unbiased coin is chosen:

P(A|E_3) = 50\% = 0.5 = \frac{1}{2}

.

To Find:

P(E_1|A)

, the probability that it was the two-headed coin given that the toss showed heads.

Solution: We use Bayes' Theorem:

P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)P(A|E_3)}

.

P(E_1|A) = \frac{\frac{1}{3} \times 1}{\left(\frac{1}{3} \times 1\right) + \left(\frac{1}{3} \times \frac{3}{4}\right) + \left(\frac{1}{3} \times \frac{1}{2}\right)}

We can cancel out the common factor of

\frac{1}{3}

from the numerator and denominator.

P(E_1|A) = \frac{1}{1 + \frac{3}{4} + \frac{1}{2}} = \frac{1}{\frac{4+3+2}{4}} = \frac{1}{\frac{9}{4}} = \frac{4}{9}

Final Answer: The probability that it was the two-headed coin is

\frac{4}{9}

.

Q7EXERCISE 13.3

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

Solution

Given: Total number of insured persons =

2000 + 4000 + 6000 = 12000

.

Let:

E_1

: The insured person is a scooter driver.

E_2

: The insured person is a car driver.

E_3

: The insured person is a truck driver. A: The insured person meets with an accident.

Probabilities of being a certain type of driver:

P(E_1) = \frac{2000}{12000} = \frac{1}{6}

P(E_2) = \frac{4000}{12000} = \frac{2}{6} = \frac{1}{3}

P(E_3) = \frac{6000}{12000} = \frac{3}{6} = \frac{1}{2}

Conditional probabilities of an accident:

P(A|E_1) = 0.01

P(A|E_2) = 0.03

P(A|E_3) = 0.15

To Find:

P(E_1|A)

, the probability that the person is a scooter driver given that he met with an accident.

Solution: We use Bayes' Theorem:

P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)P(A|E_3)}

.

Numerator:

P(E_1)P(A|E_1) = \frac{1}{6} \times 0.01 = \frac{0.01}{6}

.

Denominator:

P(A) = \left(\frac{1}{6} \times 0.01\right) + \left(\frac{1}{3} \times 0.03\right) + \left(\frac{1}{2} \times 0.15\right)

P(A) = \frac{0.01}{6} + \frac{0.03}{3} + \frac{0.15}{2}

P(A) = \frac{0.01}{6} + 0.01 + 0.075 = \frac{0.01}{6} + 0.085

P(A) = \frac{0.01 + 6 \times 0.085}{6} = \frac{0.01 + 0.51}{6} = \frac{0.52}{6}

.

Calculation:

P(E_1|A) = \frac{0.01/6}{0.52/6} = \frac{0.01}{0.52} = \frac{1}{52}

Final Answer: The probability that the person who met with an accident is a scooter driver is

\frac{1}{52}

.

Q8EXERCISE 13.3

A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B ?

Solution

Given: Let A be the event that an item is produced by machine A. Let B be the event that an item is produced by machine B. Let D be the event that the chosen item is defective.

P(A) = 60\% = 0.6

P(B) = 40\% = 0.4

Probability of a defective item given it was produced by machine A:

P(D|A) = 2\% = 0.02

.

Probability of a defective item given it was produced by machine B:

P(D|B) = 1\% = 0.01

.

To Find:

P(B|D)

, the probability that a defective item was produced by machine B.

Solution: We use Bayes' Theorem:

P(B|D) = \frac{P(B)P(D|B)}{P(A)P(D|A) + P(B)P(D|B)}

.

Numerator:

P(B)P(D|B) = 0.4 \times 0.01 = 0.004

.

Denominator (Total probability of an item being defective):

P(D) = P(A)P(D|A) + P(B)P(D|B)

P(D) = (0.6 \times 0.02) + (0.4 \times 0.01)

P(D) = 0.012 + 0.004 = 0.016

.

Calculation:

P(B|D) = \frac{0.004}{0.016} = \frac{4}{16} = \frac{1}{4}

Final Answer: The probability that the defective item was produced by machine B is

\frac{1}{4}

.

Q9EXERCISE 13.3

Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.

Solution

Given: Let

G_1

be the event that the first group wins. Let

G_2

be the event that the second group wins. Let N be the event that a new product is introduced.

P(G_1) = 0.6

P(G_2) = 0.4

Probability of introducing a new product given the first group wins:

P(N|G_1) = 0.7

.

Probability of introducing a new product given the second group wins:

P(N|G_2) = 0.3

.

To Find:

P(G_2|N)

, the probability that the second group won given that a new product was introduced.

Solution: We use Bayes' Theorem:

P(G_2|N) = \frac{P(G_2)P(N|G_2)}{P(G_1)P(N|G_1) + P(G_2)P(N|G_2)}

.

Numerator:

P(G_2)P(N|G_2) = 0.4 \times 0.3 = 0.12

.

Denominator (Total probability of a new product being introduced):

P(N) = P(G_1)P(N|G_1) + P(G_2)P(N|G_2)

P(N) = (0.6 \times 0.7) + (0.4 \times 0.3)

P(N) = 0.42 + 0.12 = 0.54

.

Calculation:

P(G_2|N) = \frac{0.12}{0.54} = \frac{12}{54} = \frac{2}{9}

Final Answer: The probability that the new product was introduced by the second group is

\frac{2}{9}

.

Q10EXERCISE 13.3

Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?

Solution

Given: Let

E_1

be the event that the girl throws a 1, 2, 3, or 4. Let

E_2

be the event that the girl throws a 5 or 6. Let A be the event that she obtained exactly one head.

Probabilities of the initial die roll:

P(E_1) = \frac{4}{6} = \frac{2}{3}

P(E_2) = \frac{2}{6} = \frac{1}{3}

Conditional probabilities of getting exactly one head: If

E_1

occurs (she throws 1, 2, 3, or 4), she tosses a coin once. The probability of getting exactly one head is:

P(A|E_1) = P(H) = \frac{1}{2}

.

If

E_2

occurs (she throws 5 or 6), she tosses a coin three times. The sample space has

2^3 = 8

outcomes. The outcomes with exactly one head are {HTT, THT, TTH}, so there are 3 such outcomes. The probability of getting exactly one head in three tosses is:

P(A|E_2) = \frac{3}{8}

.

To Find:

P(E_1|A)

, the probability that she threw 1, 2, 3, or 4, given that she obtained exactly one head.

Solution: We use Bayes' Theorem:

P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}

.

Numerator:

P(E_1)P(A|E_1) = \frac{2}{3} \times \frac{1}{2} = \frac{1}{3}

.

Denominator (Total probability of getting exactly one head):

P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2)

P(A) = \left(\frac{2}{3} \times \frac{1}{2}\right) + \left(\frac{1}{3} \times \frac{3}{8}\right)

P(A) = \frac{1}{3} + \frac{1}{8} = \frac{8+3}{24} = \frac{11}{24}

.

Calculation:

P(E_1|A) = \frac{1/3}{11/24} = \frac{1}{3} \times \frac{24}{11} = \frac{8}{11}

Final Answer: The probability that she threw 1, 2, 3 or 4 with the die is

\frac{8}{11}

.

Q11EXERCISE 13.3

A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, where as the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A ?

Solution

Given: Let A, B, C be the events that the item is produced by operator A, B, or C, respectively. Let D be the event that a defective item is produced.

Probabilities of an operator being on the job:

P(A) = 50\% = 0.5

P(B) = 30\% = 0.3

P(C) = 20\% = 0.2

Conditional probabilities of producing a defective item:

P(D|A) = 1\% = 0.01

P(D|B) = 5\% = 0.05

P(D|C) = 7\% = 0.07

To Find:

P(A|D)

, the probability that the defective item was produced by operator A.

Solution: We use Bayes' Theorem:

P(A|D) = \frac{P(A)P(D|A)}{P(A)P(D|A) + P(B)P(D|B) + P(C)P(D|C)}

.

Numerator:

P(A)P(D|A) = 0.5 \times 0.01 = 0.005

.

Denominator (Total probability of producing a defective item):

P(D) = (0.5 \times 0.01) + (0.3 \times 0.05) + (0.2 \times 0.07)

P(D) = 0.005 + 0.015 + 0.014 = 0.034

.

Calculation:

P(A|D) = \frac{0.005}{0.034} = \frac{5}{34}

Final Answer: The probability that the defective item was produced by A is

\frac{5}{34}

.

Q12EXERCISE 13.3

A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.

Solution

Given: One card is lost from a 52-card deck. Two cards are drawn from the remaining 51 cards and are both diamonds.

Let:

E_1

: The lost card is a diamond.

E_2

: The lost card is not a diamond. A: The two cards drawn from the remaining 51 cards are both diamonds.

Probabilities of the lost card: There are 13 diamonds and 39 non-diamonds in a deck.

P(E_1) = \frac{13}{52} = \frac{1}{4}

P(E_2) = \frac{39}{52} = \frac{3}{4}

Conditional probabilities of drawing two diamonds: If the lost card was a diamond (

E_1

occurred), there are 12 diamonds left in the remaining 51 cards. The probability of drawing two diamonds is:

P(A|E_1) = \frac{\text{Number of ways to choose 2 diamonds from 12}}{\text{Number of ways to choose 2 cards from 51}} = \frac{C(12, 2)}{C(51, 2)}

P(A|E_1) = \frac{12 \times 11 / 2}{51 \times 50 / 2} = \frac{132}{2550} = \frac{22}{425}

.

If the lost card was not a diamond (

E_2

occurred), there are still 13 diamonds left in the remaining 51 cards. The probability of drawing two diamonds is:

P(A|E_2) = \frac{C(13, 2)}{C(51, 2)} = \frac{13 \times 12 / 2}{51 \times 50 / 2} = \frac{156}{2550} = \frac{26}{425}

.

To Find:

P(E_1|A)

, the probability that the lost card was a diamond, given that two diamonds were drawn.

Solution: We use Bayes' Theorem:

P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}

.

P(E_1|A) = \frac{\frac{1}{4} \times \frac{22}{425}}{\left(\frac{1}{4} \times \frac{22}{425}\right) + \left(\frac{3}{4} \times \frac{26}{425}\right)}

We can cancel the common factor of

\frac{1}{4 \times 425}

from the numerator and denominator.

P(E_1|A) = \frac{22}{22 + (3 \times 26)} = \frac{22}{22 + 78} = \frac{22}{100} = \frac{11}{50}

Final Answer: The probability of the lost card being a diamond is

\frac{11}{50}

.

Q13EXERCISE 13.3

Probability that A speaks truth is $\frac{4}{5}$ . A coin is tossed. A reports that a head appears. The probability that actually there was head is

(A)

\frac{4}{5}

(B)

\frac{1}{2}

(C)

\frac{1}{5}

(D)

\frac{2}{5}

Solution

Given: Let T be the event that A speaks the truth.

P(T) = \frac{4}{5}

. Let T' be the event that A lies.

P(T') = 1 - \frac{4}{5} = \frac{1}{5}

.

Let:

E_1

: The actual outcome of the coin toss is a Head (H).

E_2

: The actual outcome of the coin toss is a Tail (T). A: A reports that a head appears.

Prior probabilities of the coin toss:

P(E_1) = \frac{1}{2}

P(E_2) = \frac{1}{2}

Conditional probabilities of A's report: If a head actually occurred (

E_1

), A reporting a head means he is speaking the truth.

P(A|E_1) = P(T) = \frac{4}{5}

. If a tail actually occurred (

E_2

), A reporting a head means he is lying.

P(A|E_2) = P(T') = \frac{1}{5}

.

To Find:

P(E_1|A)

, the probability that there was actually a head, given that A reported a head.

Solution: We use Bayes' Theorem:

P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}

.

P(E_1|A) = \frac{\frac{1}{2} \times \frac{4}{5}}{\left(\frac{1}{2} \times \frac{4}{5}\right) + \left(\frac{1}{2} \times \frac{1}{5}\right)}

We can cancel the common factor of

\frac{1}{2}

from the numerator and denominator.

P(E_1|A) = \frac{\frac{4}{5}}{\frac{4}{5} + \frac{1}{5}} = \frac{\frac{4}{5}}{\frac{5}{5}} = \frac{4/5}{1} = \frac{4}{5}

Final Answer: The correct option is (A)

\frac{4}{5}

.

Q14EXERCISE 13.3

If A and B are two events such that $A \subset B$ and $P(B) \neq 0$ , then which of the following is correct?

(A)

P(A|B) = \frac{P(B)}{P(A)}

(B)

P(A|B) < P(A)

(C)

P(A|B) \geq P(A)

(D)

None of these

Solution

Given: A and B are two events such that

A \subset B

and

P(B) \neq 0

.

To Find: The correct relationship between

P(A|B)

and

P(A)

.

Solution: The formula for conditional probability is

P(A|B) = \frac{P(A \cap B)}{P(B)}

. Since A is a subset of B (

A \subset B

), the intersection of A and B is A itself (

A \cap B = A

). Therefore,

P(A \cap B) = P(A)

.

Substituting this into the formula:

P(A|B) = \frac{P(A)}{P(B)}

Now we need to compare this with

P(A)

. We know that for any event B, its probability is between 0 and 1, i.e.,

0 \leq P(B) \leq 1

. Since we are given

P(B) \neq 0

, we have

0 < P(B) \leq 1

.

From this inequality, we can say that

\frac{1}{P(B)} \geq 1

.

Now, let's multiply both sides of this inequality by

P(A)

(which is non-negative):

P(A) \times \frac{1}{P(B)} \geq P(A) \times 1

\frac{P(A)}{P(B)} \geq P(A)

Since

P(A|B) = \frac{P(A)}{P(B)}

, we have:

P(A|B) \geq P(A)

The equality holds if

P(B)=1

. The inequality is strict if

P(B)<1

.

Let's check the options: (A)

P(A|B) = \frac{P(B)}{P(A)}

is incorrect. It should be

\frac{P(A)}{P(B)}

. (B)

P(A|B) < P(A)

is incorrect. (C)

P(A|B) \geq P(A)

is correct.

Final Answer: The correct option is (C)

P(A|B) \geq P(A)

.

Q1Miscellaneous Exercise on Chapter 13

A and B are two events such that $P(A) \neq 0$ . Find $P(B|A)$ , if

(i)

A is a subset of B

(ii)

$A \cap B = \phi$

Solution

Given: A and B are two events with

P(A) \neq 0

.

To Find:

P(B|A)

under two different conditions.

Formula: The conditional probability of B given A is

P(B|A) = \frac{P(A \cap B)}{P(A)}

.

Solution:

(i) A is a subset of B ( $A \subset B$ ) If A is a subset of B, then every outcome in A is also in B. This means the intersection of A and B is the set A itself.

A \cap B = A

. Therefore,

P(A \cap B) = P(A)

.

Now, substituting this into the formula for conditional probability:

P(B|A) = \frac{P(A)}{P(A)} = 1

(ii) $A \cap B = \phi$ If the intersection of A and B is the empty set (

\phi

), it means A and B are mutually exclusive events. They cannot happen at the same time. The probability of the empty set is 0.

P(A \cap B) = P(\phi) = 0

.

Now, substituting this into the formula for conditional probability:

P(B|A) = \frac{0}{P(A)}

Since we are given

P(A) \neq 0

, the result is:

P(B|A) = 0

Final Answer:

(i)

If A is a subset of B,

P(B|A) = 1

.

(ii)

If

A \cap B = \phi

,

P(B|A) = 0

.

Q2Miscellaneous Exercise on Chapter 13

A couple has two children,

(i)

Find the probability that both children are males, if it is known that at least one of the children is male.

(ii)

Find the probability that both children are females, if it is known that the elder child is a female.

Solution

Given: A couple has two children. Let M denote a male and F denote a female. The sample space is:

S = \{MM, MF, FM, FF\}

, where the first letter represents the elder child. Each outcome is equally likely with a probability of

\frac{1}{4}

.

Solution:

(i) Find P(both males | at least one is male) Let E be the event 'both children are males'.

E = \{MM\}

. Let A be the event 'at least one of the children is male'.

A = \{MM, MF, FM\}

.

We need to find

P(E|A)

. First, find the intersection

E \cap A

. The outcome common to E and A is {MM}.

E \cap A = \{MM\}

.

The conditional probability is given by:

P(E|A) = \frac{n(E \cap A)}{n(A)} = \frac{1}{3}

(ii) Find P(both females | elder child is a female) Let G be the event 'both children are females'.

G = \{FF\}

. Let B be the event 'the elder child is a female'.

B = \{FM, FF\}

.

We need to find

P(G|B)

. First, find the intersection

G \cap B

. The outcome common to G and B is {FF}.

G \cap B = \{FF\}

.

The conditional probability is given by:

P(G|B) = \frac{n(G \cap B)}{n(B)} = \frac{1}{2}

Final Answer:

(i)

The probability that both children are males, given that at least one is male, is

\frac{1}{3}

.

(ii)

The probability that both children are females, given that the elder child is a female, is

\frac{1}{2}

.

Q3Miscellaneous Exercise on Chapter 13

Suppose that 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.

Solution

Given: Let M be the event that the selected person is a male. Let F be the event that the selected person is a female. Let G be the event that the selected person has grey hair.

From the problem statement: There are an equal number of males and females, so:

P(M) = 0.5

P(F) = 0.5

5% of men have grey hair:

P(G|M) = 5\% = 0.05

.

0.25% of women have grey hair:

P(G|F) = 0.25\% = 0.0025

.

To Find:

P(M|G)

, the probability that the person is male, given that they have grey hair.

Solution: We use Bayes' Theorem:

P(M|G) = \frac{P(M)P(G|M)}{P(M)P(G|M) + P(F)P(G|F)}

.

Numerator:

P(M)P(G|M) = 0.5 \times 0.05 = 0.025

.

Denominator (Total probability of having grey hair):

P(G) = P(M)P(G|M) + P(F)P(G|F)

P(G) = (0.5 \times 0.05) + (0.5 \times 0.0025)

P(G) = 0.025 + 0.00125 = 0.02625

.

Calculation:

P(M|G) = \frac{0.025}{0.02625} = \frac{2500}{2625}

To simplify the fraction, divide numerator and denominator by their greatest common divisor, which is 125.

P(M|G) = \frac{2500 \div 125}{2625 \div 125} = \frac{20}{21}

Final Answer: The probability of the grey-haired person being male is

\frac{20}{21}

.

Q4Miscellaneous Exercise on Chapter 13

Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?

Solution

Given: This is a binomial distribution problem. Let X be the number of right-handed people in a sample. Sample size,

n = 10

. Probability of a person being right-handed (success),

p = 90\% = 0.9

. Probability of a person not being right-handed (failure),

q = 1 - p = 1 - 0.9 = 0.1

.

To Find: The probability that at most 6 people are right-handed, i.e.,

P(X \leq 6)

.

Formula: The probability of k successes in n trials in a binomial distribution is given by:

P(X=k) = C(n, k) p^k q^{n-k}

Solution:

P(X \leq 6) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6)

. Calculating this directly is tedious. It is easier to calculate the complement event,

P(X > 6)

, and subtract from 1.

P(X \leq 6) = 1 - P(X > 6) = 1 - [P(X=7) + P(X=8) + P(X=9) + P(X=10)]

.

Let's calculate the probabilities for X = 7, 8, 9, 10:

P(X=7) = C(10, 7) (0.9)^7 (0.1)^3 = \frac{10!}{7!3!} (0.9)^7 (0.1)^3 = 120 (0.9)^7 (0.1)^3

P(X=8) = C(10, 8) (0.9)^8 (0.1)^2 = \frac{10!}{8!2!} (0.9)^8 (0.1)^2 = 45 (0.9)^8 (0.1)^2

P(X=9) = C(10, 9) (0.9)^9 (0.1)^1 = \frac{10!}{9!1!} (0.9)^9 (0.1)^1 = 10 (0.9)^9 (0.1)^1

P(X=10) = C(10, 10) (0.9)^{10} (0.1)^0 = 1 (0.9)^{10}

This calculation is still complex without a calculator. The question likely expects the answer to be left in this form in an exam context, or it's from a section where calculators are assumed. Let's express the answer as a sum.

P(X \leq 6) = \sum_{k=0}^{6} C(10, k) (0.9)^k (0.1)^{10-k}

Final Answer: The probability that at most 6 of a random sample of 10 people are right-handed is given by the sum

\sum_{k=0}^{6} C(10, k) (0.9)^k (0.1)^{10-k}

.

Q5Miscellaneous Exercise on Chapter 13

If a leap year is selected at random, what is the chance that it will contain 53 tuesdays?

Solution

Given: A leap year is selected at random.

To Find: The probability that the leap year contains 53 Tuesdays.

Solution: A leap year has 366 days. We can find the number of full weeks in a leap year by dividing 366 by 7.

366 = 52 \times 7 + 2

This means a leap year consists of 52 complete weeks and 2 extra days.

Each of the 52 weeks will have one Tuesday, so there are guaranteed to be 52 Tuesdays. For the leap year to have 53 Tuesdays, one of the two extra days must be a Tuesday.

The possible combinations for these two consecutive extra days are:

Sunday, Monday
Monday, Tuesday
Tuesday, Wednesday
Wednesday, Thursday
Thursday, Friday
Friday, Saturday
Saturday, Sunday

There are 7 equally likely outcomes for the pair of extra days. The outcomes that are favorable to having a 53rd Tuesday are the pairs that contain 'Tuesday'. These are:

Monday, Tuesday
Tuesday, Wednesday

There are 2 favorable outcomes.

The probability is the ratio of favorable outcomes to the total number of outcomes.

P(53 \text{ Tuesdays}) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{2}{7}

Final Answer: The chance that a randomly selected leap year will contain 53 Tuesdays is

\frac{2}{7}

.

Q6Miscellaneous Exercise on Chapter 13

Suppose we have four boxes A, B, C and D containing coloured marbles as given below:
Box Red White Black
A 1 6 3
B 6 2 2
C 8 1 1
D 0 6 4
One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?, box B?, box C?

Box	Red	White	Black
A	1	6	3
B	6	2	2
C	8	1	1
D	0	6	4

Solution

Given: Four boxes with different numbers of marbles. Total marbles in each box: Box A:

1+6+3=10

Box B:

6+2+2=10

Box C:

8+1+1=10

Box D:

0+6+4=10

Let:

E_A, E_B, E_C, E_D

be the events of selecting box A, B, C, D respectively. Let R be the event that the drawn marble is red.

Since a box is selected at random:

P(E_A) = P(E_B) = P(E_C) = P(E_D) = \frac{1}{4}

.

Conditional probabilities of drawing a red marble:

P(R|E_A) = \frac{1}{10}

P(R|E_B) = \frac{6}{10}

P(R|E_C) = \frac{8}{10}

P(R|E_D) = \frac{0}{10} = 0

To Find:

P(E_A|R)

,

P(E_B|R)

, and

P(E_C|R)

.

Solution: We use Bayes' Theorem. First, let's find the total probability of drawing a red marble,

P(R)

.

P(R) = P(E_A)P(R|E_A) + P(E_B)P(R|E_B) + P(E_C)P(R|E_C) + P(E_D)P(R|E_D)

P(R) = (\frac{1}{4} \times \frac{1}{10}) + (\frac{1}{4} \times \frac{6}{10}) + (\frac{1}{4} \times \frac{8}{10}) + (\frac{1}{4} \times 0)

P(R) = \frac{1}{40} + \frac{6}{40} + \frac{8}{40} + 0 = \frac{15}{40} = \frac{3}{8}

.

Now we can find the required probabilities:

1. Probability it was drawn from box A:

P(E_A|R) = \frac{P(E_A)P(R|E_A)}{P(R)} = \frac{\frac{1}{4} \times \frac{1}{10}}{\frac{15}{40}} = \frac{\frac{1}{40}}{\frac{15}{40}} = \frac{1}{15}

2. Probability it was drawn from box B:

P(E_B|R) = \frac{P(E_B)P(R|E_B)}{P(R)} = \frac{\frac{1}{4} \times \frac{6}{10}}{\frac{15}{40}} = \frac{\frac{6}{40}}{\frac{15}{40}} = \frac{6}{15} = \frac{2}{5}

3. Probability it was drawn from box C:

P(E_C|R) = \frac{P(E_C)P(R|E_C)}{P(R)} = \frac{\frac{1}{4} \times \frac{8}{10}}{\frac{15}{40}} = \frac{\frac{8}{40}}{\frac{15}{40}} = \frac{8}{15}

Final Answer:

The probability it was drawn from box A is $\frac{1}{15}$ .
The probability it was drawn from box B is $\frac{2}{5}$ .
The probability it was drawn from box C is $\frac{8}{15}$ .

Q7Miscellaneous Exercise on Chapter 13

Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a meditation and yoga course reduce the risk of heart attack by 30% and prescription of certain drug reduces its chances by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?

Solution

Given: Let

P_0

be the initial probability of a heart attack,

P_0 = 40\% = 0.4

.

Let

E_1

be the event that the patient chooses the meditation and yoga course. Let

E_2

be the event that the patient chooses the drug prescription. Let H be the event that the patient suffers a heart attack.

Since the patient chooses an option with equal probabilities:

P(E_1) = \frac{1}{2}

P(E_2) = \frac{1}{2}

Conditional probabilities of a heart attack: Meditation and yoga reduce the risk by 30%. The new risk is

100\% - 30\% = 70\%

of the original risk.

P(H|E_1) = P_0 \times (1 - 0.30) = 0.4 \times 0.7 = 0.28

.

The drug reduces the risk by 25%. The new risk is

100\% - 25\% = 75\%

of the original risk.

P(H|E_2) = P_0 \times (1 - 0.25) = 0.4 \times 0.75 = 0.30

.

To Find:

P(E_1|H)

, the probability that the patient followed a course of meditation and yoga, given that they suffered a heart attack.

Solution: We use Bayes' Theorem:

P(E_1|H) = \frac{P(E_1)P(H|E_1)}{P(E_1)P(H|E_1) + P(E_2)P(H|E_2)}

.

Numerator:

P(E_1)P(H|E_1) = \frac{1}{2} \times 0.28 = 0.14

.

Denominator (Total probability of a heart attack under these options):

P(H) = (\frac{1}{2} \times 0.28) + (\frac{1}{2} \times 0.30)

P(H) = 0.14 + 0.15 = 0.29

.

Calculation:

P(E_1|H) = \frac{0.14}{0.29} = \frac{14}{29}

Final Answer: The probability that the patient followed a course of meditation and yoga is

\frac{14}{29}

.

Q8Miscellaneous Exercise on Chapter 13

If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability $\frac{1}{2}$ )

Solution

Given: A second order determinant is of the form

\begin{vmatrix} a & b \ c & d \end{vmatrix}

. Each element

a, b, c, d

can be either 0 or 1. The value of the determinant is

ad - bc

.

To Find: The probability that the value of the determinant is positive, i.e.,

ad - bc > 0

.

Solution: Since each of the four elements can take one of two values (0 or 1), the total number of possible determinants is

2 \times 2 \times 2 \times 2 = 2^4 = 16

. Each of these 16 determinants is equally likely.

We need to find the cases where

ad - bc > 0

. Since the elements can only be 0 or 1, the products

ad

and

bc

can only be 0 or 1. For

ad - bc

to be positive, the only possibility is:

ad = 1

and

bc = 0

.

Let's find the number of ways these conditions can be met:

Condition 1: $ad = 1$ This requires both

a=1

and

d=1

. This is only 1 case.

Condition 2: $bc = 0$ This requires that at least one of

b

or

c

is 0. The possible pairs for

(b, c)

are:

(0, 0)
(0, 1)
(1, 0) There are 3 cases for this condition.

The total number of favorable outcomes is the number of ways Condition 1 can happen multiplied by the number of ways Condition 2 can happen. Number of favorable outcomes =

1 \times 3 = 3

.

The specific determinants are:

$a=1, d=1, b=0, c=0$ : $\begin{vmatrix} 1 & 0 \ 0 & 1 \end{vmatrix} = 1 - 0 = 1$
$a=1, d=1, b=0, c=1$ : $\begin{vmatrix} 1 & 0 \ 1 & 1 \end{vmatrix} = 1 - 0 = 1$
$a=1, d=1, b=1, c=0$ : $\begin{vmatrix} 1 & 1 \ 0 & 1 \end{vmatrix} = 1 - 0 = 1$

So, there are 3 determinants with a positive value.

The probability is the ratio of favorable outcomes to the total number of outcomes.

P(\text{determinant is positive}) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{3}{16}

Final Answer: The probability that the value of the determinant is positive is

\frac{3}{16}

.

Q9Miscellaneous Exercise on Chapter 13

An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known: $P(A \text{ fails}) = 0.2$ $P(B \text{ fails alone}) = 0.15$ $P(A \text{ and } B \text{ fail}) = 0.15$ Evaluate the following probabilities

(i)

P(A fails | B has failed)

(ii)

P(A fails alone)

Solution

Given: Let

A_f

be the event that subsystem A fails. Let

B_f

be the event that subsystem B fails.

P(A_f) = 0.2

P(B \text{ fails alone}) = P(B_f \cap A_f') = 0.15

P(A \text{ and } B \text{ fail}) = P(A_f \cap B_f) = 0.15

Solution:

(i) P(A fails | B has failed) We need to find

P(A_f | B_f)

. The formula is:

P(A_f | B_f) = \frac{P(A_f \cap B_f)}{P(B_f)}

We are given

P(A_f \cap B_f) = 0.15

. We need to find

P(B_f)

.

The event 'B fails' (

B_f

) can be partitioned into two mutually exclusive events:

B fails and A fails ( $B_f \cap A_f$ ).
B fails and A does not fail ( $B_f \cap A_f'$ ), which is 'B fails alone'. So, $P(B_f) = P(B_f \cap A_f) + P(B_f \cap A_f')$ . $P(B_f) = 0.15 + 0.15 = 0.30$ .

Now we can calculate the conditional probability:

P(A_f | B_f) = \frac{0.15}{0.30} = \frac{1}{2} = 0.5

(ii) P(A fails alone) We need to find the probability that A fails and B does not fail, which is

P(A_f \cap B_f')

.

The event 'A fails' (

A_f

) can be partitioned into two mutually exclusive events:

A fails and B fails ( $A_f \cap B_f$ ).
A fails and B does not fail ( $A_f \cap B_f'$ ), which is 'A fails alone'. So, $P(A_f) = P(A_f \cap B_f) + P(A_f \cap B_f')$ . $0.2 = 0.15 + P(A_f \cap B_f')$ .

Rearranging the equation:

P(A_f \cap B_f') = 0.2 - 0.15 = 0.05

.

Final Answer:

(i)

P(A fails | B has failed) = 0.5

(ii)

P(A fails alone) = 0.05

Q10Miscellaneous Exercise on Chapter 13

Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

Solution

Given: Bag I: 3 Red (R), 4 Black (B). Total = 7. Bag II: 4 Red (R), 5 Black (B). Total = 9.

Let:

E_1

: The ball transferred from Bag I to Bag II is red.

E_2

: The ball transferred from Bag I to Bag II is black. A: The ball drawn from Bag II is red.

Probabilities of the transferred ball:

P(E_1) = \frac{3}{7}

(Probability of drawing a red ball from Bag I).

P(E_2) = \frac{4}{7}

(Probability of drawing a black ball from Bag I).

Conditional probabilities of drawing a red ball from Bag II: If a red ball was transferred (

E_1

occurred), Bag II now has

4+1=5

Red and 5 Black balls. Total = 10.

P(A|E_1) = \frac{5}{10} = \frac{1}{2}

.

If a black ball was transferred (

E_2

occurred), Bag II now has 4 Red and

5+1=6

Black balls. Total = 10.

P(A|E_2) = \frac{4}{10} = \frac{2}{5}

.

To Find:

P(E_2|A)

, the probability that the transferred ball was black, given that the ball drawn from Bag II is red.

Solution: We use Bayes' Theorem:

P(E_2|A) = \frac{P(E_2)P(A|E_2)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}

.

Numerator:

P(E_2)P(A|E_2) = \frac{4}{7} \times \frac{2}{5} = \frac{8}{35}

.

Denominator (Total probability of drawing a red ball from Bag II):

P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2)

P(A) = \left(\frac{3}{7} \times \frac{1}{2}\right) + \left(\frac{4}{7} \times \frac{2}{5}\right)

P(A) = \frac{3}{14} + \frac{8}{35} = \frac{15 + 16}{70} = \frac{31}{70}

.

Calculation:

P(E_2|A) = \frac{8/35}{31/70} = \frac{8}{35} \times \frac{70}{31} = \frac{8 \times 2}{31} = \frac{16}{31}

Final Answer: The probability that the transferred ball is black is

\frac{16}{31}

.

Q11Miscellaneous Exercise on Chapter 13

If A and B are two events such that $P(A) \neq 0$ and $P(B|A) = 1$ , then

(A)

A \subset B

(B)

B \subset A

(C)

B = \phi

(D)

A = \phi

Solution

Given: A and B are two events such that

P(A) \neq 0

and

P(B|A) = 1

.

Solution: The definition of conditional probability is

P(B|A) = \frac{P(A \cap B)}{P(A)}

. We are given that

P(B|A) = 1

. So,

1 = \frac{P(A \cap B)}{P(A)}

Multiplying both sides by

P(A)

(which is non-zero), we get:

P(A \cap B) = P(A)

The probability of the intersection of two events can be equal to the probability of one of the events only if that event is a subset of the other. In this case,

P(A \cap B) = P(A)

implies that the event A is a subset of the event B. This means that whenever event A occurs, event B must also occur. Therefore,

A \subset B

.

Analysis of options: (A)

A \subset B

: This is consistent with our conclusion. (B)

B \subset A

: This would imply

P(A \cap B) = P(B)

, which is not what we derived. (C)

B = \phi

: If

B = \phi

, then

A \cap B = \phi

, so

P(A \cap B) = 0

. This would make

P(B|A) = 0

, contradicting the given information. (D)

A = \phi

: This would mean

P(A) = 0

, which contradicts the given information

P(A) \neq 0

.

Final Answer: The correct option is (A)

A \subset B

.

Q12Miscellaneous Exercise on Chapter 13

If $P(A|B) > P(A)$ , then which of the following is correct:

(A)

P(B|A) < P(B)

(B)

P(A \cap B) < P(A) \cdot P(B)

(C)

P(B|A) > P(B)

(D)

P(B|A) = P(B)

Solution

Given:

P(A|B) > P(A)

. We assume

P(A)

and

P(B)

are non-zero.

Solution: Start with the given inequality:

P(A|B) > P(A)

Using the definition of conditional probability, substitute

P(A|B) = \frac{P(A \cap B)}{P(B)}

:

\frac{P(A \cap B)}{P(B)} > P(A)

Multiply both sides by

P(B)

(which is positive):

P(A \cap B) > P(A) \cdot P(B)

This inequality tells us that the occurrence of A and B together is more likely than if they were independent. This suggests a positive correlation between the events.

Now, let's analyze the expression for

P(B|A)

:

P(B|A) = \frac{P(B \cap A)}{P(A)}

Since

A \cap B = B \cap A

, we have

P(A \cap B) = P(B \cap A)

. So,

P(B|A) = \frac{P(A \cap B)}{P(A)}

.

From our derived inequality, we know

P(A \cap B) > P(A) \cdot P(B)

. Substitute this into the expression for

P(B|A)

:

P(B|A) > \frac{P(A) \cdot P(B)}{P(A)}

Since

P(A) > 0

, we can cancel it out:

P(B|A) > P(B)

Analysis of options: (A)

P(B|A) < P(B)

: Incorrect. (B)

P(A \cap B) < P(A) \cdot P(B)

: Incorrect. We found the opposite is true. (C)

P(B|A) > P(B)

: Correct. (D)

P(B|A) = P(B)

: This would imply independence, which contradicts the given inequality.

Final Answer: The correct option is (C)

P(B|A) > P(B)

.

Q13Miscellaneous Exercise on Chapter 13

If A and B are any two events such that $P(A) + P(B) - P(A \text{ and } B) = P(A)$ , then

(A)

P(B|A) = 1

(B)

P(A|B) = 1

(C)

P(B|A) = 0

(D)

P(A|B) = 0

Solution

Given: A and B are two events such that

P(A) + P(B) - P(A \text{ and } B) = P(A)

. Note that

P(A \text{ and } B)

is the same as

P(A \cap B)

.

Solution: Start with the given equation:

P(A) + P(B) - P(A \cap B) = P(A)

Subtract

P(A)

from both sides:

P(B) - P(A \cap B) = 0

P(B) = P(A \cap B)

This equality means that the probability of event B is the same as the probability of the intersection of A and B. This can only be true if the event B is a subset of the event A (

B \subset A

). This is because if B occurs, A must also occur for their intersection to have the same probability as B.

Now let's evaluate the options, assuming

P(B) \neq 0

so that

P(A|B)

is defined.

(A) $P(B|A)$

P(B|A) = \frac{P(A \cap B)}{P(A)}

. Since

P(A \cap B) = P(B)

, this becomes

P(B|A) = \frac{P(B)}{P(A)}

. This is not necessarily 1.

(B) $P(A|B)$

P(A|B) = \frac{P(A \cap B)}{P(B)}

. Since we derived

P(A \cap B) = P(B)

, we can substitute this into the formula:

P(A|B) = \frac{P(B)}{P(B)} = 1

(C) $P(B|A) = 0$ This would require

P(A \cap B) = 0

, which means

P(B) = 0

. While possible, it's not the general conclusion.

(D) $P(A|B) = 0$ This contradicts our finding that

P(A|B) = 1

.

Final Answer: The correct option is (B)

P(A|B) = 1

.

ProbabilityClass 12 Mathematics NCERT Solutions

Given that E and F are events such that P(E)=0.6P(E) = 0.6P(E)=0.6, P(F)=0.3P(F) = 0.3P(F)=0.3 and P(E∩F)=0.2P(E \cap F) = 0.2P(E∩F)=0.2, find P(E∣F)P(E|F)P(E∣F) and P(F∣E)P(F|E)P(F∣E)

Solution

Compute P(A∣B)P(A|B)P(A∣B), if P(B)=0.5P(B) = 0.5P(B)=0.5 and P(A∩B)=0.32P(A \cap B) = 0.32P(A∩B)=0.32

Solution

If P(A)=0.8P(A) = 0.8P(A)=0.8, P(B)=0.5P(B) = 0.5P(B)=0.5 and P(B∣A)=0.4P(B|A) = 0.4P(B∣A)=0.4, find

P(A∩B)P(A \cap B)P(A∩B)

P(A∣B)P(A|B)P(A∣B)

P(A∪B)P(A \cup B)P(A∪B)

Solution

Evaluate P(A∪B)P(A \cup B)P(A∪B), if 2P(A)=P(B)=5132P(A) = P(B) = \frac{5}{13}2P(A)=P(B)=135​ and P(A∣B)=25P(A|B) = \frac{2}{5}P(A∣B)=52​

Solution

If P(A)=611P(A) = \frac{6}{11}P(A)=116​, P(B)=511P(B) = \frac{5}{11}P(B)=115​ and P(A∪B)=711P(A \cup B) = \frac{7}{11}P(A∪B)=117​, find

P(A∩B)P(A \cap B)P(A∩B)

P(A∣B)P(A|B)P(A∣B)

P(B∣A)P(B|A)P(B∣A)

Solution

A coin is tossed three times, where

E : head on third toss, F : heads on first two tosses

E : at least two heads, F : at most two heads

E : at most two tails, F : at least one tail Determine P(E∣F)P(E|F)P(E∣F) in each case.

Solution

Two coins are tossed once, where

E : tail appears on one coin, F : one coin shows head

E : no tail appears, F : no head appears Determine P(E∣F)P(E|F)P(E∣F) in each case.

Solution

A die is thrown three times, E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses Determine P(E∣F)P(E|F)P(E∣F).

Solution

Mother, father and son line up at random for a family picture E : son on one end, F : father in middle Determine P(E∣F)P(E|F)P(E∣F).

Solution

A black and a red dice are rolled.

Solution

A fair die is rolled. Consider events E={1,3,5}E = \{1,3,5\}E={1,3,5}, F={2,3}F = \{2,3\}F={2,3} and G={2,3,4,5}G = \{2,3,4,5\}G={2,3,4,5}. Find

P(E∣F)P(E|F)P(E∣F) and P(F∣E)P(F|E)P(F∣E)

P(E∣G)P(E|G)P(E∣G) and P(G∣E)P(G|E)P(G∣E)

P((E∪F)∣G)P((E \cup F)|G)P((E∪F)∣G) and P((E∩F)∣G)P((E \cap F)|G)P((E∩F)∣G)

Solution

Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl, (ii) at least one is a girl?

Solution

Solution

Given that the two numbers appearing on throwing two dice are different. Find the probability of the event 'the sum of numbers on the dice is 4'.

Solution

Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event 'the coin shows a tail', given that 'at least one die shows a 3'.

Solution

If P(A)=12P(A) = \frac{1}{2}P(A)=21​, P(B)=0P(B) = 0P(B)=0, then P(A∣B)P(A|B)P(A∣B) is

Solution

If A and B are events such that P(A∣B)=P(B∣A)P(A|B) = P(B|A)P(A∣B)=P(B∣A), then

Solution

If P(A)=35P(A) = \frac{3}{5}P(A)=53​ and P(B)=15P(B) = \frac{1}{5}P(B)=51​, find P(A∩B)P(A \cap B)P(A∩B) if A and B are independent events.

Solution

Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.

Solution

Solution

A fair coin and an unbiased die are tossed. Let A be the event 'head appears on the coin' and B be the event '3 on the die'. Check whether A and B are independent events or not.

Solution

A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, 'the number is even,' and B be the event, 'the number is red'. Are A and B independent?

Solution

Let E and F be events with P(E)=35P(E) = \frac{3}{5}P(E)=53​, P(F)=310P(F) = \frac{3}{10}P(F)=103​ and P(E∩F)=15P(E \cap F) = \frac{1}{5}P(E∩F)=51​. Are E and F independent?

Solution

Given that the events A and B are such that P(A)=12P(A) = \frac{1}{2}P(A)=21​, P(A∪B)=35P(A \cup B) = \frac{3}{5}P(A∪B)=53​ and P(B)=pP(B) = pP(B)=p. Find ppp if they are (i) mutually exclusive (ii) independent.

Solution

Let A and B be independent events with P(A)=0.3P(A) = 0.3P(A)=0.3 and P(B)=0.4P(B) = 0.4P(B)=0.4. Find

P(A∩B)P(A \cap B)P(A∩B)

P(A∪B)P(A \cup B)P(A∪B)

P(A∣B)P(A|B)P(A∣B)

P(B∣A)P(B|A)P(B∣A)

Solution

If A and B are two events such that P(A)=14P(A) = \frac{1}{4}P(A)=41​, P(B)=12P(B) = \frac{1}{2}P(B)=21​ and P(A∩B)=18P(A \cap B) = \frac{1}{8}P(A∩B)=81​, find P(not A and not B).

Solution

Events A and B are such that P(A)=12P(A) = \frac{1}{2}P(A)=21​, P(B)=712P(B) = \frac{7}{12}P(B)=127​ and P(not A or not B) = 14\frac{1}{4}41​. State whether A and B are independent ?

Solution

Given two independent events A and B such that P(A)=0.3P(A) = 0.3P(A)=0.3, P(B)=0.6P(B) = 0.6P(B)=0.6. Find

P(A and B)P(A \text{ and } B)P(A and B)

P(A and not B)P(A \text{ and not } B)P(A and not B)

P(A or B)P(A \text{ or } B)P(A or B)

P(neither A nor B)

Solution

A die is tossed thrice. Find the probability of getting an odd number at least once.

Solution

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that

Given that E and F are events such that $P(E) = 0.6$ , $P(F) = 0.3$ and $P(E \cap F) = 0.2$ , find $P(E|F)$ and $P(F|E)$

Compute $P(A|B)$ , if $P(B) = 0.5$ and $P(A \cap B) = 0.32$

If $P(A) = 0.8$ , $P(B) = 0.5$ and $P(B|A) = 0.4$ , find

$P(A \cap B)$

$P(A|B)$

$P(A \cup B)$

Evaluate $P(A \cup B)$ , if $2P(A) = P(B) = \frac{5}{13}$ and $P(A|B) = \frac{2}{5}$

If $P(A) = \frac{6}{11}$ , $P(B) = \frac{5}{11}$ and $P(A \cup B) = \frac{7}{11}$ , find

$P(A \cap B)$

$P(A|B)$

$P(B|A)$

E : at most two tails, F : at least one tail Determine $P(E|F)$ in each case.

E : no tail appears, F : no head appears Determine $P(E|F)$ in each case.

A die is thrown three times, E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses Determine $P(E|F)$ .

Mother, father and son line up at random for a family picture E : son on one end, F : father in middle Determine $P(E|F)$ .

A fair die is rolled. Consider events $E = \{1,3,5\}$ , $F = \{2,3\}$ and $G = \{2,3,4,5\}$ . Find

$P(E|F)$ and $P(F|E)$

$P(E|G)$ and $P(G|E)$

$P((E \cup F)|G)$ and $P((E \cap F)|G)$

If $P(A) = \frac{1}{2}$ , $P(B) = 0$ , then $P(A|B)$ is

If A and B are events such that $P(A|B) = P(B|A)$ , then

If $P(A) = \frac{3}{5}$ and $P(B) = \frac{1}{5}$ , find $P(A \cap B)$ if A and B are independent events.

Let E and F be events with $P(E) = \frac{3}{5}$ , $P(F) = \frac{3}{10}$ and $P(E \cap F) = \frac{1}{5}$ . Are E and F independent?

Given that the events A and B are such that $P(A) = \frac{1}{2}$ , $P(A \cup B) = \frac{3}{5}$ and $P(B) = p$ . Find $p$ if they are (i) mutually exclusive (ii) independent.

Let A and B be independent events with $P(A) = 0.3$ and $P(B) = 0.4$ . Find

$P(A \cap B)$

$P(A \cup B)$

$P(A|B)$

$P(B|A)$

If A and B are two events such that $P(A) = \frac{1}{4}$ , $P(B) = \frac{1}{2}$ and $P(A \cap B) = \frac{1}{8}$ , find P(not A and not B).

Events A and B are such that $P(A) = \frac{1}{2}$ , $P(B) = \frac{7}{12}$ and P(not A or not B) = $\frac{1}{4}$ . State whether A and B are independent ?

Given two independent events A and B such that $P(A) = 0.3$ , $P(B) = 0.6$ . Find

$P(A \text{ and } B)$

$P(A \text{ and not } B)$

$P(A \text{ or } B)$

Probability of solving specific problem independently by A and B are $\frac{1}{2}$ and $\frac{1}{3}$ respectively. If both try to solve the problem independently, find the probability that

Probability that A speaks truth is $\frac{4}{5}$ . A coin is tossed. A reports that a head appears. The probability that actually there was head is

If A and B are two events such that $A \subset B$ and $P(B) \neq 0$ , then which of the following is correct?

A and B are two events such that $P(A) \neq 0$ . Find $P(B|A)$ , if

$A \cap B = \phi$

If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability $\frac{1}{2}$ )