Welcome! If you’re looking for the NCERT Solutions for Class 12 Maths Chapter 13, Exercise 13.1, you’re in the right place. Below, you will find detailed, step-by-step solutions for every question in Exercise 13.1, designed to help you clear all your doubts and understand the logic behind each answer.
This exercise introduces the core concepts of conditional probability, which are important for your board exams.
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).
Solution:
Let’s solve this by breaking it into the two parts the question asks for.
Part 1: Finding P(E|F)
P(E|F). This notation stands for the conditional probability of event E, given that event F has already occurred.
Formula for conditional probability is: P(E|F) = \frac{P(E \cap F)}{P(F)}
From the question, we have:
P(E \cap F) = 0.2
P(F) = 0.3
Substitute and Solve: Now, we just plug these values into our formula:
P(E|F) = \frac{0.2}{0.3}
P(E|F) = \frac{2}{3}
Part 2: Finding P(F|E)
P(F|E). This is the probability of event F, given that event E has already occurred. Notice the order is reversed, which is very important!
The formula is similar, but this time, the “given” event is E, so P(E) goes in the denominator:
P(F|E) = \frac{P(F \cap E)}{P(E)}
From the question, we have:
P(E \cap F) = 0.2 (which is the same as P(F \cap E))
P(E) = 0.6
Substitute and Solve: Let’s plug these values into the formula for P(F|E):
P(F|E) = \frac{0.2}{0.6}
P(F|E) = \frac{2}{6}
P(F|E) = \frac{1}{3}
Final Answer: \boxed{ P(E|F) = \frac{2}{3} \ \text{ and } \ P(F|E) = \frac{1}{3} }
Q2. Compute P(A|B) if P(B) = 0.5 and P(A \cap B) = 0.32
Solution:
We need to find P(A|B), which is the conditional probability of A given that B has occurred.
The formula for conditional probability is: P(A|B) = \frac{P(A \cap B)}{P(B)}
From the question, we have:
P(A \cap B) = 0.32
P(B) = 0.5
Substitute and Solve: Now, we plug these values into our formula:
P(A|B) = \frac{0.32}{0.5}
P(A|B) = \frac{32}{50}
P(A|B) = \frac{16}{25}
Final Answer: \boxed{ P(A|B) = \frac{16}{25} }
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).
Solution:
(i): Finding P(A \cap B)
We are given P(B|A).
The formula for this is P(B|A) = \frac{P(A \cap B)}{P(A)}.
We can rearrange this formula to find P(A \cap B)
P(A \cap B) = P(B|A) \times P(A) \quad … (1)
From the question, we have:
P(B|A) = 0.4
P(A) = 0.8
Substitute in (1) and Solve:
P(A \cap B) = 0.4 \times 0.8 = 0.32
(ii): Finding P(A|B)
Now we need to find P(A|B).
The formula is: P(A|B) = \frac{P(A \cap B)}{P(B)} … (2)
From the question and Part (i), we have:
P(A \cap B) = 0.32
P(B) = 0.5
Substitute in (2) and Solve:
P(A|B) = \frac{0.32}{0.5} = \frac{32}{50} = 0.64
(iii): Finding P(A \cup B)
We know: P(A \cup B) = P(A) + P(B) - P(A \cap B)
From the question and Part (i), we have:
P(A) = 0.8
P(B) = 0.5
P(A \cap B) = 0.32
Substitute and Solve:
P(A \cup B) = 0.8 + 0.5 - 0.32
P(A \cup B) = 1.3 - 0.32 = 0.98
Final Answer: \boxed{ (i) P(A \cap B) = 0.32, (ii) P(A|B) = 0.64, (iii) P(A \cup B) = 0.98 }
Q4. Evaluate P(A \cup B) if 2P(A) = P(B) = \frac{5}{13} and P(A|B) = \frac{2}{5}.
Solution:
This problem requires us to first find all the individual probabilities needed for the addition rule.
Addition Rule: P(A \cup B) = P(A) + P(B) - P(A \cap B) … (1)
Find P(A) and P(B)
We are given P(B) = \frac{5}{13}.
We are also given 2P(A) = \frac{5}{13}.
Solving for P(A), we get P(A) = \frac{1}{2} \times \frac{5}{13} = \frac{5}{26}.
Find P(A \cap B)
We use the multiplication rule derived from conditional probability: P(A \cap B) = P(A|B) \times P(B) … (2)
From the question and we have:
P(A|B) = \frac{2}{5}
P(B) = \frac{5}{13}
Substitute in (2)
P(A \cap B) = \frac{2}{5} \times \frac{5}{13} = \frac{2}{13}
Find P(A \cup B)
From our previous steps, we have:
P(A) = \frac{5}{26}
P(B) = \frac{5}{13}
P(A \cap B) = \frac{2}{13}
Substitute in (1)
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: \boxed{ P(A \cup B) = \frac{11}{26} }
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).
Solution:
(i). Finding P(A \cap B)
We know P(A \cup B) = P(A) + P(B) - P(A \cap B).
P(A \cap B) = P(A) + P(B) - P(A \cup B) \quad … (1)
We have:
P(A) = \frac{6}{11}
P(B) = \frac{5}{11}
P(A \cup B) = \frac{7}{11}
Substitute in (1)
P(A \cap B) = \frac{6}{11} + \frac{5}{11} - \frac{7}{11}
P(A \cap B) = \frac{11 - 7}{11} = \frac{4}{11}
(ii). Finding P(A|B)
The formula is: P(A|B) = \frac{P(A \cap B)}{P(B)} \quad … (2)
We have:
P(A \cap B) = \frac{4}{11}
P(B) = \frac{5}{11}
Substitute in (2) :
P(A|B) = \frac{4/11}{5/11} = \frac{4}{5}
(iii). Finding P(B|A)
The formula is: P(B|A) = \frac{P(A \cap B)}{P(A)} \quad … (3)
From the question and Part (i), we have:
P(A \cap B) = \frac{4}{11}
P(A) = \frac{6}{11}
Substitute in (3) and Solve:
P(B|A) = \frac{4/11}{6/11} = \frac{4}{6} = \frac{2}{3}
Final Answer: \boxed{ (i) P(A \cap B) = \frac{4}{11}, (ii) P(A|B) = \frac{4}{5}, (iii) P(B|A) = \frac{2}{3} }
Q6. A coin is tossed three times. Determine P(E|F) where:
(i) E: head on third toss, F: heads on first two tosses 13
(ii) E: at least two heads, F: at most two heads 14
(iii) E: at most two tails, F: at least one tail 15
Solution:
First, let’s define the sample space (S) for tossing a coin three times. The total outcomes are 2^3 = 8 .
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
(i): E: head on third toss, F: heads on first two tosses
Identify the outcomes for E and F:
E = {HH\underline{H}, HT\underline{H}, TH\underline{H}, TT\underline{H}}
F = {\underline{HH}H, \underline{HH}T}
E \cap F = {HHH}
We know: P(E|F) = \frac{n(E \cap F)}{n(F)}
P(E|F) = \frac{1}{2}
(ii): E: at least two heads, F: at most two heads
Identify the outcomes for E and F:
E = (at least 2 heads, so 2 or 3): {HHH, HHT, HTH, THH}
F = (at most 2 heads, so 0, 1, or 2): {HHT, HTH, THH, HTT, THT, TTH, TTT}
E \cap F = {HHT, HTH, THH}
We know P(E|F) = \frac{n(E \cap F)}{n(F)}
P(E|F) = \frac{3}{7}
(iii): E: at most two tails, F: at least one tail
Identify the outcomes for E and F:
E = (at most 2 tails, so 0, 1, or 2): {HHH, HHT, HTH, THH, HTT, THT, TTH}
F = (at least 1 tail, so 1, 2, or 3): {HHT, HTH, THH, HTT, THT, TTH, TTT}
E \cap F = {HHT, HTH, THH, HTT, THT, TTH}
We know P(E|F) = \frac{n(E \cap F)}{n(F)}:
P(E|F) = \frac{6}{7}
Final Answer: \boxed{ (i) P(E|F) = \frac{1}{2}, (ii) P(E|F) = \frac{3}{7}, (iii) P(E|F) = \frac{6}{7} }
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
Solution:
Let’s define the sample space (S) for tossing two coins. The total number of outcomes are 2^2 = 4 .
S = {HH, HT, TH, TT}
(i): E: tail appears on one coin, F: one coin shows head
The wording “on one coin” implies exactly one.
E = (exactly one tail): {HT, TH}
F = (exactly one head): {HT, TH}
E \cap F = {HT, TH}
We know P(E|F) = \frac{n(E \cap F)}{n(F)}
P(E|F) = \frac{2}{2} = 1
(ii): E: no tail appears, F: no head appears
Identify the outcomes for E and F:
E = (no tail): {HH}
F = (no head): {TT}
E \cap F = \emptyset (the empty set)
We know: P(E|F) = \frac{n(E \cap F)}{n(F)}:
P(E|F) = \frac{0}{1} = 0
Final Answer: \boxed{ (i) P(E|F) = 1, (ii) P(E|F) = 0 }
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).
Solution:
A die is thrown three times. The total number of possible outcomes is 6 \times 6 \times 6 = 216
Identify E, F, and E \cap F
E = “4 appears on the third toss”.
Outcomes are of the form (x, y, 4), where x and y can be any of 6 numbers.
E = (1,1,4), (1,2,4), ..., (1,6,4)
\quad (2,1,4), (2,2,4), ..., (2,6,4)
\quad (3,1,4), (3,2,4), ..., (3,6,4)
\quad (4,1,4), (4,2,4), ..., (4,6,4)
\quad (5,1,4), (5,2,4), ..., (5,6,4)
\quad (6,1,4), (6,2,4), ..., (6,6,4)
\implies n(E) = 36
F = “6 and 5 appears respectively on first two tosses”.
Outcomes are of the form (6, 5, z), where z can be any of 6 numbers.
F = {(6,5,1), (6,5,2), (6,5,3), (6,5,4), (6,5,5), (6,5,6)}
\implies n(F)= 6
E \cap F = This is the event where both F happens AND E happens.
This means the outcome must be (6, 5, z) and (x, y, 4).
The only outcome that satisfies both is (6, 5, 4).
E \cap F = {(6,5,4)}
The formula is: P(E|F) = \frac{n(E \cap F)}{n(F)}
P(E|F) = \frac{1}{6}
Final Answer: \boxed{ P(E|F) = \frac{1}{6} }
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).
Solution:
Let M = Mother, F = Father, S = Son.
Identify E, F, and E \cap F
E = “son on one end” (Son is first or Son is last)
E = {\underline{S}MF, \underline{S}FM, MF\underline{S}, FM\underline{S}}
F = “father in middle”
F = {M\underline{F}S, S\underline{F}M}
E \cap F = The outcomes that are in both E and F.
E \cap F = {MFS, SFM}
P(E \cap F) = \frac{n(E \cap F)}{n(F)} = \frac{2}{2}
P(E|F) = 1
Final Answer: \boxed{ P(E|F) = 1 }
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.
(b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.</h4>
Solution:
(a) Sum greater than 9, given black die is 5
Let A be the event ‘the sum is greater than 9’.
Let F be the event ‘the black die resulted in a 5’.
We need to find P(A|F).
And we know: P(A|F) = \frac{n(A \cap F)}{n(F)}
Let’s find the outcomes for each event:
(Black die, Red die)
F = (Black die is 5) = {(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)}
n(F) = 6
A = (Sum > 9) = {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)}
n(A) = 6
A \cap F = (Outcomes in both A and F) = {(5,5), (5,6)}
n(A \cap F) = 2
P(A|F) = \frac{n(A \cap F)}{n(F)} = \frac{2}{6}
P(A|F) = \frac{1}{3}
(b) Sum is 8, given red die is less than 4
Let E be the event ‘the sum is 8’.
Let G be the event ‘the red die resulted in a number less than 4’.
We need to find P(E|G).
The formula is: P(E|G) = \frac{P(E \cap G)}{P(G)}
Let’s find the outcomes for each event:
G = (Red die is 1, 2, or 3) = 18 outcomes
E = (Sum is 8) = {(2,6), (3,5), (4,4), (5,3), (6,2)}
E \cap G = (Sum is 8 AND Red die is < 4)
From event E: (2,6) $\to$ Red=6 (No). (3,5) $\to$ Red=5 (No). (4,4) $\to$ Red=4 (No). (5,3) $\to$ Red=3 (Yes). (6,2) $\to$ Red=2 (Yes).
E \cap G = {(5,3), (6,2)}Now, we find the probabilities:
Substitute and Solve:
P(E|G) = \frac{2/36}{18/36} = \frac{2}{18} = \frac{1}{9}Final Answer: \boxed{ (a) P(A|F) = \frac{1}{3}, (b) P(E|G) = \frac{1}{9} }
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)
(iii) P((E \cup F)|G) and P((E \cap F)|G)
Solution:
The sample space (S) for a fair die roll is S = {1, 2, 3, 4, 5, 6}.
Calculating the probabilities of E, F, G and their intersections:
P(E) = \frac{3}{6} = \frac{1}{2}
P(F) = \frac{2}{6} = \frac{1}{3}
P(G) = \frac{4}{6} = \frac{2}{3}
And the intersections:
E \cap F = {3} \ \implies \ P(E \cap F) = \frac{1}{6}
E \cap G = {3, 5} \ \implies \ P(E \cap G) = \frac{2}{6} = \frac{1}{3}
E \cup F = {1, 2, 3, 5} \ \implies \ (E \cup F) \cap G = {2, 3, 5} \implies P((E \cup F) \cap G) = \frac{3}{6} = \frac{1}{2}
E \cap F = {3} \ \implies \ (E \cap F) \cap G = {3} \ \implies \ P((E \cap F) \cap G) = \frac{1}{6}
(i) Find P(E|F) and P(F|E)
P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{1/6}{2/6} = \frac{1}{2}
P(F|E) = \frac{P(E \cap F)}{P(E)} = \frac{1/6}{3/6} = \frac{1}{3}
(ii) Find P(E|G) and P(G|E)
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(E \cap G)}{P(E)} = \frac{2/6}{3/6} = \frac{2}{3}
(iii) Find P((E \cup F)|G) and P((E \cap F)|G)
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:
\boxed{ \begin{aligned} (i) P(E|F) = \frac{1}{2}, P(F|E) = \frac{1}{3} \end{aligned} }
\boxed{ \begin{aligned} (ii) P(E|G) = \frac{1}{2}, P(G|E) = \frac{2}{3} \end{aligned} }
\boxed{ \begin{aligned} (iii) P((E \cup F)|G) = \frac{3}{4}, P((E \cap F)|G) = \frac{1}{4} \end{aligned} }
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 girls given that
(i) the youngest is a girl
(ii) at least one is a girl?
Solution:
The sample space (S) for a family with two children is S = {BB, BG, GB, GG}, where the first letter represents the elder child and the second represents the younger. [/latex]n(S)=4 [/latex].
Let E be the event ‘both children are girls’.
E = {GG}
(i). Given that the youngest is a girl
Let F be the event: the youngest is a girl.
F = {BG, GG}
We need to find P(E|F).
E \cap F = {GG}
P(E|F) = \frac{n(E \cap F)}{n(F)} = \frac{1}{2}
(ii) Given that at least one is a girl
Let G be the event ‘at least one is a girl’.
G = {BG, GB, GG}
We need to find P(E|G)
E \cap G = {GG}
P(E|G) = \frac{n(E \cap G)}{n(G)} = \frac{1}{3}
Final Answer: \boxed{ (i) P(E|F) = \frac{1}{2}, (ii) P(E|G) = \frac{1}{3} }
Q13. 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:
Let’s find the total number of questions.
Total = 300 (Easy T/F) + 200 (Diff T/F) + 500 (Easy MCQ) + 400 (Diff MCQ) = 1400 questions.
Let E be the event ‘the question is easy’.
Let F be the event ‘the question is a multiple choice question (MCQ)’.
n(F) = (Total number of MCQs) = 500 (Easy MCQ) + 400 (Diff MCQ) = 900
E \cap F \ implies Questions that are Easy and MCQ
n(E \cap F) = (Number of questions that are Easy AND MCQ) = 500
The formula is P(E|F) = \frac{n(E \cap F)}{n(F)}.
P(E|F) = \frac{500}{900} = \frac{5}{9}
Final Answer: \boxed{ P(E|F) = \frac{5}{9} }
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’.
Solution:
When two dice are thrown, the total number of outcomes is n(S) = 36
Let E be the event ‘the sum of numbers is 4’.
E = {(1,3), (2,2), (3,1)}
n(E) = 3
Let F be the event ‘the two numbers appearing are different’.
F = (Numbers are different).
The opposite of F is ‘numbers are the same’ (doubles).
Doubles = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}, which has 6 outcomes.
So, n(F) = n(S) - n(\text{doubles}) = 36 - 6 = 30.
E \cap F \implies \ Sum of the numbers is 4 and number are not different
E \cap F = {(1,3), (3,1)}
n(E \cap F) = 2
The formula for P(E|F) = \frac{n(E \cap F)}{n(F)} = \frac{n(E \cap F)}{n(F)}
The formula for P(E|F) = \frac{2}{20} = \frac{1}{15}
Final Answer: \boxed{ P(E|F) = \frac{1}{15} }
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 conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.
Solution:
This experiment has a non-uniform sample space. This means that the probability of all the outcomes is not same.
Case 1: First roll is {1, 2, 4, 5}. (Prob = \frac{4}{6}). Then toss a coin (H or T).
Case 2: First roll is {3, 6}. (Prob = \frac{2}{6}). Then roll a die (1, 2, 3, 4, 5, 6).
Sample Space S = (1,H), (1, T), (2, H), (2, T), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, H), (4, T), (5, H), (5, T), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
Let E be the event ‘the coin shows a tail’.
E = {(1,T), (2,T), (4,T), (5,T)}
Let F be the event ‘at least one die shows a 3’.
F = {(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (6,3)}
n(F) = 6
E \cap F = Empty Set
n (E \cap F) = 0
We know P(E|F) = \frac{n(E \cap F)}{n(F)}
P(E|F) = \frac{0}{6}
Final Answer: \boxed{ P(E|F) = 0 }
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
Solution:
The formula for conditional probability is P(A|B) = \frac{P(A \cap B)}{P(B)} … (1)
Given:
P(B) = 0
Substituting in (1)
P(A|B) = \frac{P(A \cap B)}{0}
Therefore, P(A|B) is not defined.
Correct Option: (C)
Q17. If A and B are events such that P(A|B) = P(B|A), then
(A) A \subset B \text{ but } A \neq B
(B) A = B
(C) A \cap B = \emptyset
(D) P(A) = P(B)
Solution:
Given P(A|B) = P(B|A).
Let’s assume P(A) \neq 0 and P(B) \neq 0, otherwise the probabilities are not defined.
Using the formula of conditional probability
\frac{P(A \cap B)}{P(B)} = \frac{P(B \cap A)}{P(A)}
This gives us: \frac{1}{P(B)} = \frac{1}{P(A)}
P(A) = P(B)
Correct Option: (D)