Exercise 13.1, Question 1
\(P(E) = 0.6\), \(P(F) = 0.3\), \(P(E \cap F) = 0.2\). Find \(P(E|F)\) and \(P(F|E)\).
Show solution
- \(P(E|F) = \dfrac{0.2}{0.3}\), \(P(F|E) = \dfrac{0.2}{0.6}\).
Answer: \(P(E|F) = \tfrac23\), \(P(F|E) = \tfrac13\)
NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.1
Exercise 13.1: Conditional probability. \(P(E|F) = \dfrac{P(E \cap F)}{P(F)}\) (\(P(F) \ne 0\)): restrict the sample space to F. With equally likely outcomes, \(P(E|F) = \dfrac{n(E \cap F)}{n(F)}\). \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).
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Exercise 13.1 questions and solutions
Exercise 13.1, Question 2
Find \(P(A|B)\) if \(P(B) = 0.5\) and \(P(A \cap B) = 0.32\).
Show solution
- \(\dfrac{0.32}{0.5}\).
Answer: \(0.64\)
Exercise 13.1, Question 3
\(P(A) = 0.8\), \(P(B) = 0.5\), \(P(B|A) = 0.4\). Find:
(i) \(P(A \cap B)\)
Show solution
- \(P(A)P(B|A) = 0.8 \times 0.4\).
Answer: \(0.32\)
(ii) \(P(A|B)\)
Show solution
- \(\dfrac{0.32}{0.5}\).
Answer: \(0.64\)
(iii) \(P(A \cup B)\)
Show solution
- \(0.8 + 0.5 - 0.32\).
Answer: \(0.98\)
Exercise 13.1, Question 4
Find \(P(A \cup B)\) if \(2P(A) = P(B) = \tfrac{5}{13}\) and \(P(A|B) = \tfrac25\).
Show solution
- \(P(A) = \tfrac{5}{26}\); \[\begin{aligned}P(A \cap B) &= P(A|B)P(B) \\ &= \tfrac25 \cdot \tfrac{5}{13} \\ &= \tfrac{2}{13}\end{aligned}\]
- \[P(A \cup B) = \tfrac{5}{26} + \tfrac{10}{26} - \tfrac{4}{26}\]
Answer: \(\tfrac{11}{26}\)
Exercise 13.1, Question 5
\(P(A) = \tfrac{6}{11}\), \(P(B) = \tfrac{5}{11}\), \(P(A \cup B) = \tfrac{7}{11}\). Find:
(i) \(P(A \cap B)\)
Show solution
- \[\tfrac{6}{11} + \tfrac{5}{11} - \tfrac{7}{11}\]
Answer: \(\tfrac{4}{11}\)
(ii) \(P(A|B)\)
Show solution
- \(\dfrac{4/11}{5/11}\).
Answer: \(\tfrac45\)
(iii) \(P(B|A)\)
Show solution
- \(\dfrac{4/11}{6/11}\).
Answer: \(\tfrac23\)
Exercise 13.1, Question 6
A coin is tossed three times. Find \(P(E|F)\) for:
(i) E: head on the third toss; F: heads on the first two tosses
Show solution
- \(F = \{HHH, HHT\}\), \(E \cap F = \{HHH\}\).
Answer: \(\tfrac12\)
(ii) E: at least two heads; F: at most two heads
Show solution
- F has 7 outcomes (all but HHH); \(E \cap F = \{HHT, HTH, THH\}\).
Answer: \(\tfrac37\)
(iii) E: at most two tails; F: at least one tail
Show solution
- F: all but HHH (7 outcomes); \(E \cap F\): all but HHH and TTT (6 outcomes).
Answer: \(\tfrac67\)
Exercise 13.1, Question 7
Two coins are tossed once. Find \(P(E|F)\) for:
(i) E: a tail on one coin (exactly one); F: one coin shows a head
Show solution
- \(E = F = \{HT, TH\}\).
Answer: 1
(ii) E: no tail; F: no head
Show solution
- \(E = \{HH\}\), \(F = \{TT\}\), \(E \cap F = \emptyset\).
Answer: 0
Exercise 13.1, Question 8
A die is thrown three times. E: 4 on the third toss; F: 6 and 5 on the first two tosses, in that order. Find \(P(E|F)\).
Show solution
- F has 6 outcomes \((6, 5, k)\); one of them has \(k = 4\).
Answer: \(\tfrac16\)
Exercise 13.1, Question 9
Mother, father and son line up at random. E: son at one end; F: father in the middle. Find \(P(E|F)\).
Show solution
- F: MFS, SFM. In both, the son is at an end.
Answer: 1
Exercise 13.1, Question 10
A black and a red die are rolled.
(a) Find the probability of a sum greater than 9, given the black die shows 5.
Show solution
- Given black = 5, the red die must show 5 or 6.
Answer: \(\tfrac13\)
(b) Find the probability of the sum 8, given the red die shows a number less than 4.
Show solution
- 18 outcomes have red \(\in \{1, 2, 3\}\); sum 8 needs black 6 with red 2, or black 5 with red 3.
Answer: \(\tfrac19\)
Exercise 13.1, Question 11
A fair die is rolled; \(E = \{1, 3, 5\}\), \(F = \{2, 3\}\), \(G = \{2, 3, 4, 5\}\). Find:
(i) \(P(E|F)\) and \(P(F|E)\)
Show solution
- \(E \cap F = \{3\}\): \(\tfrac{1}{2}\) and \(\tfrac13\).
Answer: \(\tfrac12,\ \tfrac13\)
(ii) \(P(E|G)\) and \(P(G|E)\)
Show solution
- \(E \cap G = \{3, 5\}\): \(\tfrac24\) and \(\tfrac23\).
Answer: \(\tfrac12,\ \tfrac23\)
(iii) \(P((E \cup F)|G)\) and \(P((E \cap F)|G)\)
Show solution
- \(E \cup F = \{1, 2, 3, 5\}\), and with G: \(\{2, 3, 5\}\): \(\tfrac34\).
- \(E \cap F = \{3\}\), with G: \(\{3\}\): \(\tfrac14\).
Answer: \(\tfrac34,\ \tfrac14\)
Exercise 13.1, Question 12
A family has two children, each equally likely to be a boy or a girl. Find the probability that both are girls given that:
(i) the youngest is a girl
Show solution
- Outcomes (elder, younger) with younger G: BG, GG.
Answer: \(\tfrac12\)
(ii) at least one is a girl
Show solution
- BG, GB, GG.
Answer: \(\tfrac13\)
Exercise 13.1, Question 13
A bank has 300 easy and 200 difficult true/false questions, and 500 easy and 400 difficult multiple choice questions. Find the probability that a random question is easy, given that it is multiple choice.
Show solution
- \(\dfrac{500}{500 + 400}\).
Answer: \(\tfrac59\)
Exercise 13.1, Question 14
Two dice show different numbers. Find the probability that the sum is 4.
Show solution
- 30 outcomes have different numbers; sum 4 with different numbers: \((1, 3), (3, 1)\).
Answer: \(\tfrac{1}{15}\)
Exercise 13.1, Question 15
A die is thrown; on a multiple of 3 it is thrown again, otherwise a coin is tossed. Find the probability that the coin shows a tail, given that at least one die shows a 3.
Show solution
- If a die shows 3, the first throw was 3 or 6 and the second step was a die, not a coin.
- So 'coin shows tail' and 'at least one die shows 3' cannot happen together.
Answer: 0
Exercise 13.1, Question 16
If \(P(A) = \tfrac12\), \(P(B) = 0\), then \(P(A|B)\) is: (A) 0 (B) \(\tfrac12\) (C) not defined (D) 1
Show solution
- \(P(A|B)\) needs \(P(B) \ne 0\).
Answer: (C)
Exercise 13.1, Question 17
If \(P(A|B) = P(B|A)\), then: (A) \(A \subset B\), \(A \ne B\) (B) \(A = B\) (C) \(A \cap B = \emptyset\) (D) \(P(A) = P(B)\)
Show solution
- \[\dfrac{P(A \cap B)}{P(B)} = \dfrac{P(A \cap B)}{P(A)}\], so (when \(P(A \cap B) \ne 0\)) \(P(A) = P(B)\).
Answer: (D)
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