Question 1

Q1: 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 $P(F|E)$

Accepted Answer

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 $P(B) 
eq 0$. Solution: 1. To find $P(E|F)$: $$P(E|F) = \frac{P(E \cap F)}{P(F)}$$ $$P(E|F) = \frac{0.2}{0.3} = \frac{2}{3}$$ 2. To fin...

Question 2

Q10: 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. (...

Accepted Answer

Given: A black and a red die are rolled. The sample space S has $6 	imes 6 = 36$ 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 bla...

Question 3

Q11: 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 $P(G|E)$...

Accepted Answer

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 = \{...

Question 4

Q12: 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 g...

Accepted Answer

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...

Question 5

Q13: An instructor has a question bank consisting of 300 easy True / False questions, 200 difficult True / False questions, 500 easy multiple choice questi...

Accepted Answer

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)...

Question 6

Q14: 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'.

Accepted Answer

Given: Two dice are thrown. The total number of outcomes in the sample space is $6 	imes 6 = 36$. 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 w...

Question 7

Q15: 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 condi...

Accepted Answer

Given: An experiment is performed as follows: 1. A die is thrown. Let the outcome be $d_1$. 2. 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 $(d_1, d_2)$. 3. If $d_1 \in \{1, 2, 4, 5\}$, a coin is tossed. Let the outcome be $c$. The...

Question 8

Q16: 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

Accepted Answer

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. Thi...

Question 9

Q17: If A and B are events such that $P(A|B) = P(B|A)$, then (A) $A \subset B$ but $A 
eq B$ (B) $A = B$ (C) $A \cap B = \phi$ (D) $P(A) = P(B)$

Accepted Answer

Given: A and B are events such that $P(A|B) = P(B|A)$. We assume $P(A) 
eq 0$ and $P(B) 
eq 0$ 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)...

Question 10

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

Accepted Answer

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 $P(B) 
eq 0$. 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: $P(A|B) = 0.64$

Question 11

Q3: 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) $P(A \cup B)$

Accepted Answer

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: $$P(A \cap B) = P(B|A) 	imes P(A)$$ $$P(A \cap B) = 0.4 	imes 0.8 = 0.32$$ (ii) To find $P(A|B...

Question 12

Q4: Evaluate $P(A \cup B)$, if $2P(A) = P(B) = \frac{5}{13}$ and $P(A|B) = \frac{2}{5}$

Accepted Answer

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)...

Question 13

Q5: 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) $P(B|A)$

Accepted Answer

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} + \...

Question 14

Q6: 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)...

Accepted Answer

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\...

Question 15

Q7: 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(...

Accepted Answer

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...

Question 16

Q8: A die is thrown three times, E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses Determine $P(E|F)$.

Accepted Answer

Given: A die is thrown three times. The total number of outcomes in the sample space is $6 	imes 6 	imes 6 = 216$. 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, $n(E) = 6 	imes 6 	imes 1 = 36$. Eve...

Question 17

Q9: Mother, father and son line up at random for a family picture E : son on one end, F : father in middle Determine $P(E|F)$.

Accepted Answer

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 posi...

Question 18

Q1: If $P(A) = \frac{3}{5}$ and $P(B) = \frac{1}{5}$, find $P(A \cap B)$ if A and B are independent events.

Accepted Answer

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: $$P(A \cap B) = P(A) 	imes P(B)$$ Solution: $$P(A \cap B) = \frac{3}{5} 	imes \frac{1}{5}$$ $$P(A \cap...

Question 19

Q10: 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...

Accepted Answer

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 $P(A \cap B) = P(A) 	imes P(B)$. Solution: First, we need to find $P(A \cap B)$ from the given inf...

Question 20

Q11: Given two independent events A and B such that $P(A) = 0.3$, $P(B) = 0.6$. Find (i) $P(A 	ext{ and } B)$ (ii) $P(A 	ext{ and not } B)$ (iii) $P(A 	...

Accepted Answer

Given: A and B are independent events. $P(A) = 0.3$ $P(B) = 0.6$ Solution: (i) $P(A 	ext{ and } B)$ This is $P(A \cap B)$. Since A and B are independent: $$P(A \cap B) = P(A) 	imes P(B) = 0.3 	imes 0.6 = 0.18$$ (ii) $P(A 	ext{ and not } B)$ This is $P(A \cap B')$. If A and B are independent, the...

Box	Red	White	Black
A	1	6	3
B	6	2	2
C	8	1	1
D	0	6	4

NCERT Solutions