ProbabilityClass 11 Mathematics NCERT Solutions

Given: A die is rolled. The sample space is $S = \{1, 2, 3, 4, 5, 6\}$. Event E: "die shows 4". So, $E = \{4\}$. Event F: "die shows even number". So, $F = \{2, 4, 6\}$.

To determine: Whether events E and F are mutually exclusive.

Solution: Two events are mutually exclusive if they cannot occur at the same time, which means their intersection is an empty set. We find the intersection of E and F: $E \cap F = \{4\} \cap \{2, 4, 6\} = \{4\}$ Since the intersection $E \cap F$ is not the empty set ($\phi$), the events can occur simultaneously (if the die shows 4).

Final Answer: No, E and F are not mutually exclusive because $E \cap F = \{4\} \neq \phi$.

Given: An experiment of rolling a pair of dice. The sample space S consists of 36 ordered pairs $(x, y)$ where $x, y \in \{1, 2, 3, 4, 5, 6\}$.

Checking for Mutually Exclusive Pairs: Two events are mutually exclusive if their intersection is the empty set ($\phi$).

Final Answer: The pairs of mutually exclusive events are (A, B) and (B, C).

Given: Three coins are tossed once. The sample space is: $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$

(i) Mutually Exclusive Events: We check pairs of events for an empty intersection.

(ii) Simple Events: A simple event has only one sample point.

(iii) Compound Events: A compound event has more than one sample point.

Given: Three coins are tossed. The sample space is: $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$

(i) Two events which are mutually exclusive. Let A be the event 'getting all heads' and B be the event 'getting all tails'. $A = \{HHH\}$ $B = \{TTT\}$ $A \cap B = \phi$. Thus, A and B are mutually exclusive.

(ii) Three events which are mutually exclusive and exhaustive. Let A be the event 'getting no heads', B be the event 'getting exactly one head', and C be the event 'getting at least two heads'. $A = \{TTT\}$ $B = \{HTT, THT, TTH\}$ $C = \{HHT, HTH, THH, HHH\}$

(iii) Two events, which are not mutually exclusive. Let A be the event 'getting at least two heads' and B be the event 'the first coin shows a head'. $A = \{HHH, HHT, HTH, THH\}$ $B = \{HHH, HHT, HTH, HTT\}$ $A \cap B = \{HHH, HHT, HTH\} \neq \phi$. Thus, A and B are not mutually exclusive.

(iv) Two events which are mutually exclusive but not exhaustive. Let A be the event 'getting all heads' and B be the event 'getting all tails'. $A = \{HHH\}$ $B = \{TTT\}$

(v) Three events which are mutually exclusive but not exhaustive. Let A be the event 'getting all heads', B be the event 'getting all tails', and C be the event 'getting exactly one head'. $A = \{HHH\}$ $B = \{TTT\}$ $C = \{HTT, THT, TTH\}$

Given: Two dice are thrown. The sample space S has 36 outcomes.

(i) A' (not A): This is the event that the first die does not show an even number, which means it shows an odd number. Thus, $A' = B$. $A' = \{(1,1), (1,2), ..., (1,6), (3,1), ..., (3,6), (5,1), ..., (5,6)\}$

(ii) not B (B'): This is the event that the first die does not show an odd number, which means it shows an even number. Thus, $B' = A$. $B' = \{(2,1), (2,2), ..., (2,6), (4,1), ..., (4,6), (6,1), ..., (6,6)\}$

(iii) A or B ($A \cup B$): This is the event of getting an even number or an odd number on the first die. This covers all possible outcomes. $A \cup B = S$. This is a sure event.

(iv) A and B ($A \cap B$): This is the event of getting an even number and an odd number on the first die simultaneously, which is impossible. $A \cap B = \phi$. This is an impossible event.

(v) A but not C ($A - C$ or $A \cap C'$): Getting an even number on the first die and the sum is greater than 5. From event A, we remove the elements that are also in C. Elements of C with an even first die are: $(2,1), (2,2), (2,3), (4,1)$. $A - C = A - \{(2,1), (2,2), (2,3), (4,1)\}$ This set contains all pairs from A except these four.

(vi) B or C ($B \cup C$): Getting an odd number on the first die, or the sum is $\leq 5$. $B \cup C = \{(1,1), ..., (1,6), (3,1), ..., (3,6), (5,1), ..., (5,6)\} \cup \{(2,1), (2,2), (2,3), (4,1)\}$

(vii) B and C ($B \cap C$): Getting an odd number on the first die and the sum is $\leq 5$. $B \cap C = \{(1,1), (1,2), (1,3), (1,4), (3,1), (3,2), (5,1)\}$ -- Wait, (5,1) has sum 6, so it's not in C. Let's recheck C. C is correct. (5,1) is in B but not C. So the intersection is: $B \cap C = \{(1,1), (1,2), (1,3), (1,4), (3,1), (3,2)\}$

(viii) $A \cap B' \cap C'$: This is 'A and not B and not C'. We know that $B' = A$. So, $A \cap B' = A \cap A = A$. The event becomes $A \cap C'$, which is the same as $A - C$ from part (v). So, $A \cap B' \cap C' = A - C$. The description is 'getting an even number on the first die and the sum of the numbers on the dice is greater than 5'.

Given: Events A, B, and C as defined in question 6. $A$: even number on the first die. $B$: odd number on the first die. $C$: sum of numbers $\leq 5$.

(i) A and B are mutually exclusive Answer: True. Reason: The event 'getting an even number on the first die' and 'getting an odd number on the first die' cannot happen at the same time. Mathematically, $A \cap B = \phi$.

(ii) A and B are mutually exclusive and exhaustive Answer: True. Reason: We already know they are mutually exclusive ($A \cap B = \phi$). For them to be exhaustive, their union must be the entire sample space S. $A \cup B$ represents the event that the first die is either even or odd, which is always true. Thus, $A \cup B = S$. Since both conditions are met, the statement is true.

(iii) A = B' Answer: True. Reason: The event B' (not B) means 'not getting an odd number on the first die', which is equivalent to 'getting an even number on the first die'. This is precisely the definition of event A. Therefore, $A = B'$.

(iv) A and C are mutually exclusive Answer: False. Reason: For A and C to be mutually exclusive, their intersection must be empty. However, there are outcomes where the first die is even and the sum is $\leq 5$. For example, the outcome (2, 2) is in both A and C. $A \cap C = \{(2,1), (2,2), (2,3), (4,1)\} \neq \phi$.

(v) A and B' are mutually exclusive. Answer: False. Reason: From part (iii), we know that $A = B'$. The intersection of a set with itself is the set itself. $A \cap B' = A \cap A = A$. Since A is not an empty set, they are not mutually exclusive.

(vi) A', B', C are mutually exclusive and exhaustive. Answer: False. Reason: Let's check the conditions.

Conditions for a valid probability assignment: For a sample space $S = \{\omega_1, \omega_2, ..., \omega_n\}$, a valid probability assignment must satisfy two conditions:

We will check each assignment against these conditions.

Final Answer: The assignments that cannot be valid are (c), (d), and (e).

Given: A coin is tossed twice.

To Find: The probability that at least one tail occurs.

Solution: The sample space S for tossing a coin twice is: $S = \{HH, HT, TH, TT\}$ Total number of outcomes, $n(S) = 4$.

Let E be the event that 'at least one tail occurs'. This means we can have one tail or two tails. The outcomes favorable to E are: $E = \{HT, TH, TT\}$ Number of favorable outcomes, $n(E) = 3$.

The probability of event E is given by: $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{n(E)}{n(S)}$ $P(E) = \frac{3}{4}$

Alternative Method (using complement): Let E be the event 'at least one tail occurs'. Then E' is the event 'no tail occurs', which means 'all heads occur'. $E' = \{HH\}$ $n(E') = 1$ $P(E') = \frac{n(E')}{n(S)} = \frac{1}{4}$ We know that $P(E) = 1 - P(E')$. $P(E) = 1 - \frac{1}{4} = \frac{3}{4}$

Final Answer: The probability that at least one tail occurs is $\frac{3}{4}$.

Given: A die is thrown. The sample space is $S = \{1, 2, 3, 4, 5, 6\}$. Total number of outcomes, $n(S) = 6$.

(i) A prime number will appear Let A be the event 'a prime number will appear'. The prime numbers in S are 2, 3, 5. $A = \{2, 3, 5\}$ $n(A) = 3$ $P(A) = \frac{n(A)}{n(S)} = \frac{3}{6} = \frac{1}{2}$

(ii) A number greater than or equal to 3 will appear Let B be the event 'a number greater than or equal to 3 will appear'. $B = \{3, 4, 5, 6\}$ $n(B) = 4$ $P(B) = \frac{n(B)}{n(S)} = \frac{4}{6} = \frac{2}{3}$

(iii) A number less than or equal to one will appear Let C be the event 'a number less than or equal to one will appear'. $C = \{1\}$ $n(C) = 1$ $P(C) = \frac{n(C)}{n(S)} = \frac{1}{6}$

(iv) A number more than 6 will appear Let D be the event 'a number more than 6 will appear'. There is no number in S that is greater than 6. $D = \phi$ (impossible event) $n(D) = 0$ $P(D) = \frac{n(D)}{n(S)} = \frac{0}{6} = 0$

(v) A number less than 6 will appear Let E be the event 'a number less than 6 will appear'. $E = \{1, 2, 3, 4, 5\}$ $n(E) = 5$ $P(E) = \frac{n(E)}{n(S)} = \frac{5}{6}$

Given: A card is selected from a standard pack of 52 cards.

(a) How many points are there in the sample space? The sample space consists of all 52 cards. Therefore, the total number of points (outcomes) in the sample space is 52. $n(S) = 52$.

(b) Calculate the probability that the card is an ace of spades. Let A be the event 'the card is an ace of spades'. There is only one ace of spades in a deck. Number of favorable outcomes, $n(A) = 1$. $P(A) = \frac{n(A)}{n(S)} = \frac{1}{52}$

(i) an ace Let B be the event 'the card is an ace'. There are 4 aces in a deck (spades, hearts, diamonds, clubs). Number of favorable outcomes, $n(B) = 4$. $P(B) = \frac{n(B)}{n(S)} = \frac{4}{52} = \frac{1}{13}$

(ii) black card Let C be the event 'the card is a black card'. A deck has two black suits: spades and clubs. Each suit has 13 cards. Total number of black cards = $13 + 13 = 26$. Number of favorable outcomes, $n(C) = 26$. $P(C) = \frac{n(C)}{n(S)} = \frac{26}{52} = \frac{1}{2}$

Given: A fair coin with faces marked {1, 6} and a fair die with faces marked {1, 2, 3, 4, 5, 6} are tossed.

Sample Space: The sample space S is the set of all possible pairs of outcomes (coin, die). $S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$ Total number of outcomes, $n(S) = 2 \times 6 = 12$.

(i) The sum of numbers is 3 Let A be the event that the sum of the numbers is 3. We need to find pairs in S that sum to 3.

(ii) The sum of numbers is 12 Let B be the event that the sum of the numbers is 12. We need to find pairs in S that sum to 12.

Given: Number of men = 4 Number of women = 6 Total number of council members = $4 + 6 = 10$.

To Find: The probability that the selected member is a woman.

Solution: One council member is selected at random. Total number of possible outcomes (people who can be selected), $n(S) = 10$.

Let E be the event that the selected member is a woman. Number of outcomes favorable to E (number of women), $n(E) = 6$.

The probability of event E is: $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{n(E)}{n(S)}$ $P(E) = \frac{6}{10} = \frac{3}{5}$

Final Answer: The probability that the selected council member is a woman is $\frac{3}{5}$.

Given: A fair coin is tossed four times. Win Re 1 for a head (H), lose Rs 1.50 for a tail (T). Total number of outcomes in the sample space is $2^4 = 16$.

Calculating Different Amounts: Let 'h' be the number of heads and 't' be the number of tails. We always have $h + t = 4$. Amount = $(h \times 1) - (t \times 1.50) = h - 1.5t$ Since $t = 4 - h$, Amount = $h - 1.5(4 - h) = h - 6 + 1.5h = 2.5h - 6$.

We can have the following cases for the number of heads (h):

Summary of Results: There are 5 different possible amounts of money.

Final Answer: There are 5 different amounts of money you can have:

Given: Three coins are tossed once. The sample space is: $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$ Total number of outcomes, $n(S) = 8$.

(i) 3 heads Event A = {HHH}. $n(A) = 1$. $P(A) = \frac{1}{8}$.

(ii) 2 heads Event B = {HHT, HTH, THH}. $n(B) = 3$. $P(B) = \frac{3}{8}$.

(iii) atleast 2 heads (2 or 3 heads) Event C = {HHT, HTH, THH, HHH}. $n(C) = 4$. $P(C) = \frac{4}{8} = \frac{1}{2}$.

(iv) atmost 2 heads (0, 1, or 2 heads) Event D = {HHT, HTH, THH, HTT, THT, TTH, TTT}. $n(D) = 7$. $P(D) = \frac{7}{8}$. (Alternatively, this is the complement of getting 3 heads: $1 - P(3 \text{ heads}) = 1 - \frac{1}{8} = \frac{7}{8}$)

(v) no head (same as 3 tails) Event E = {TTT}. $n(E) = 1$. $P(E) = \frac{1}{8}$.

(vi) 3 tails (same as no head) Event F = {TTT}. $n(F) = 1$. $P(F) = \frac{1}{8}$.

(vii) exactly two tails (same as exactly one head) Event G = {HTT, THT, TTH}. $n(G) = 3$. $P(G) = \frac{3}{8}$.

(viii) no tail (same as 3 heads) Event H = {HHH}. $n(H) = 1$. $P(H) = \frac{1}{8}$.

(ix) atmost two tails (0, 1, or 2 tails) Event I = {HHH, HHT, HTH, THH, HTT, THT, TTH}. $n(I) = 7$. $P(I) = \frac{7}{8}$. (Alternatively, this is the complement of getting 3 tails: $1 - P(3 \text{ tails}) = 1 - \frac{1}{8} = \frac{7}{8}$)

Given: The probability of an event A is $P(A) = \frac{2}{11}$.

To Find: The probability of the event 'not A', which is denoted as $P(A')$.

Formula: The probability of the complement of an event A is given by: $P(A') = 1 - P(A)$

Solution: Substituting the given value of P(A) into the formula: $P(A') = 1 - \frac{2}{11}$ $P(A') = \frac{11}{11} - \frac{2}{11}$ $P(A') = \frac{11 - 2}{11}$ $P(A') = \frac{9}{11}$

Final Answer: The probability of the event 'not A' is $\frac{9}{11}$.

Given: Six different natural numbers are chosen from 1 to 20.

To Find: The probability of winning the prize.

Solution: Since the order of the numbers is not important, we use combinations.

Total number of outcomes: The total number of ways to choose 6 different numbers from 20 is given by the combination formula $^{n}C_r = \frac{n!}{r!(n-r)!}$. Total number of possible combinations = $^{20}C_6$. $^{20}C_6 = \frac{20!}{6!(20-6)!} = \frac{20!}{6!14!}$ $= \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15}{6 \times 5 \times 4 \times 3 \times 2 \times 1}$ $= 19 \times 17 \times 2 \times 3 \times 5 \times 2 = 38760$ So, the total number of outcomes in the sample space is $n(S) = 38760$.

Number of favorable outcomes: To win the prize, the person's chosen six numbers must exactly match the six numbers fixed by the lottery committee. There is only one such winning combination. Number of favorable outcomes, $n(E) = 1$.

Probability of winning: $P(\text{winning}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{n(E)}{n(S)}$ $P(\text{winning}) = \frac{1}{38760}$

Final Answer: The probability of winning the prize in the game is $\frac{1}{38760}$.

Formula: The relationship between the probabilities of two events A and B is given by the addition rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

(i) Given: $P(A) = \frac{1}{3}, P(B) = \frac{1}{5}, P(A \cap B) = \frac{1}{15}$ To Find: $P(A \cup B)$ Solution: $P(A \cup B) = \frac{1}{3} + \frac{1}{5} - \frac{1}{15}$ $= \frac{5}{15} + \frac{3}{15} - \frac{1}{15}$ $= \frac{5 + 3 - 1}{15} = \frac{7}{15}$ Answer for (i): $\frac{7}{15}$

(ii) Given: $P(A) = 0.35, P(A \cap B) = 0.25, P(A \cup B) = 0.6$ To Find: $P(B)$ Solution: $0.6 = 0.35 + P(B) - 0.25$ $0.6 = 0.10 + P(B)$ $P(B) = 0.6 - 0.10 = 0.5$ We must also check that this is a valid probability. $P(A \cap B) = 0.25 \leq P(B) = 0.5$, which is true. Answer for (ii): 0.5

(iii) Given: $P(A) = 0.5, P(B) = 0.35, P(A \cup B) = 0.7$ To Find: $P(A \cap B)$ Solution: $0.7 = 0.5 + 0.35 - P(A \cap B)$ $0.7 = 0.85 - P(A \cap B)$ $P(A \cap B) = 0.85 - 0.7 = 0.15$ We must also check that this is a valid probability. $P(A \cap B) = 0.15 \leq P(A) = 0.5$ and $P(A \cap B) = 0.15 \leq P(B) = 0.35$, which are true. Answer for (iii): 0.15

Given: $P(A) = \frac{3}{5}$ $P(B) = \frac{1}{5}$ A and B are mutually exclusive events.

To Find: $P(A \text{ or } B)$, which is $P(A \cup B)$.

Formula: For any two events A and B, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. If A and B are mutually exclusive, it means they cannot occur together, so $A \cap B = \phi$ and $P(A \cap B) = 0$. The formula simplifies to: $P(A \cup B) = P(A) + P(B)$