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NCERT Solutions · Class 12 · Chapter 13: Probability

NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.3

Exercise 13.3: Total probability and Bayes' theorem. If \(E_1, \ldots, E_n\) partition the sample space: \(P(A) = \sum P(E_i)P(A|E_i)\) (total probability) and \(P(E_i|A) = \dfrac{P(E_i)P(A|E_i)}{\sum_j P(E_j)P(A|E_j)}\) (Bayes). Name the events, write the priors \(P(E_i)\) and the likelihoods \(P(A|E_i)\), then substitute; a tree diagram keeps the products straight.

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Exercise 13.3 questions and solutions

Exercise 13.3, Question 1

An urn has 5 red and 5 black balls. A ball is drawn, returned with 2 more of its colour, and a second ball is drawn. Find the probability the second ball is red.
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  1. First red (\(\tfrac12\)): urn becomes 7 red, 5 black. First black (\(\tfrac12\)): 5 red, 7 black.
  2. \[\tfrac12 \cdot \tfrac{7}{12} + \tfrac12 \cdot \tfrac{5}{12}\]
Answer: \(\tfrac12\)

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Exercise 13.3, Question 2

Bag I: 4 red, 4 black; bag II: 2 red, 6 black. A bag is chosen at random and a red ball drawn. Find the probability it came from bag I.
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  1. \[\dfrac{\frac12 \cdot \frac12}{\frac12 \cdot \frac12 + \frac12 \cdot \frac14}\]
Answer: \(\tfrac23\)

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Exercise 13.3, Question 3

60% of students are hostellers, 40% day scholars; 30% of hostellers and 20% of day scholars get A grade. A student with A grade is chosen. Find the probability they are a hosteller.
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  1. \[\dfrac{0.6 \times 0.3}{0.6 \times 0.3 + 0.4 \times 0.2} = \dfrac{0.18}{0.26}\]
Answer: \(\tfrac{9}{13}\)

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Exercise 13.3, Question 4

A student knows the answer with probability \(\tfrac34\) and guesses with probability \(\tfrac14\); a guess is right with probability \(\tfrac14\). Given the answer is correct, find the probability that the student knew it.
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  1. \[\dfrac{\frac34 \cdot 1}{\frac34 \cdot 1 + \frac14 \cdot \frac14} = \dfrac{12/16}{13/16}\]
Answer: \(\tfrac{12}{13}\)

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Exercise 13.3, Question 5

A test detects a disease 99% of the time when present, and gives a false positive for 0.5% of healthy people. 0.1% of people have the disease. Find the probability that a person with a positive result has the disease.
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  1. Let D be 'has the disease' and T be 'tests positive': \(P(D) = \tfrac{1}{1000}\), \(P(T|D) = \tfrac{99}{100}\), \(P(D') = \tfrac{999}{1000}\), \(P(T|D') = \tfrac{5}{1000}\).
  2. \[P(D|T) = \dfrac{\frac{1}{1000} \times \frac{99}{100}}{\frac{1}{1000} \times \frac{99}{100} + \frac{999}{1000} \times \frac{5}{1000}}\]; multiply top and bottom by \(10^6\): \[\dfrac{990}{990 + 4995} = \dfrac{990}{5985}\], and dividing by 45 gives \(\dfrac{22}{133}\).
Answer: \(\tfrac{22}{133}\)

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Exercise 13.3, Question 6

Three coins: two-headed, biased (heads 75% of the time), fair. One is chosen at random and shows heads. Find the probability it is the two-headed coin.
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  1. \[\dfrac{\frac13 \cdot 1}{\frac13\left(1 + \frac34 + \frac12\right)} = \dfrac{1}{9/4}\]
Answer: \(\tfrac49\)

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Exercise 13.3, Question 7

2000 scooter, 4000 car and 6000 truck drivers are insured, with accident probabilities 0.01, 0.03, 0.15. An insured person has an accident. Find the probability they drive a scooter.
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  1. \[\dfrac{2000 \times 0.01}{2000 \times 0.01 + 4000 \times 0.03 + 6000 \times 0.15} = \dfrac{20}{20 + 120 + 900}\]
Answer: \(\tfrac{1}{52}\)

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Exercise 13.3, Question 8

Machine A makes 60% of items (2% defective), machine B 40% (1% defective). A random item is defective. Find the probability it came from B.
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  1. \[\dfrac{0.4 \times 0.01}{0.6 \times 0.02 + 0.4 \times 0.01} = \dfrac{0.004}{0.016}\]
Answer: \(\tfrac14\)

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Exercise 13.3, Question 9

Two groups win the board with probabilities 0.6 and 0.4; a new product is introduced with probability 0.7 or 0.3 respectively. The product was introduced. Find the probability the second group won.
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  1. \[\dfrac{0.4 \times 0.3}{0.6 \times 0.7 + 0.4 \times 0.3} = \dfrac{0.12}{0.54}\]
Answer: \(\tfrac29\)

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Exercise 13.3, Question 10

A girl throws a die: on 5 or 6 she tosses a coin three times and notes the heads; on 1 to 4 she tosses once. She got exactly one head. Find the probability she threw 1, 2, 3 or 4.
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  1. P(exactly one head in three tosses) \(= \tfrac38\); in one toss \(= \tfrac12\).
  2. \[\dfrac{\frac23 \cdot \frac12}{\frac23 \cdot \frac12 + \frac13 \cdot \frac38} = \dfrac{1/3}{1/3 + 1/8}\]
Answer: \(\tfrac{8}{11}\)

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Exercise 13.3, Question 11

Operators A, B, C work 50%, 30%, 20% of the time with defect rates 1%, 5%, 7%. A defective item is produced. Find the probability A produced it.
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  1. \[\dfrac{0.5 \times 0.01}{0.5 \times 0.01 + 0.3 \times 0.05 + 0.2 \times 0.07} = \dfrac{0.005}{0.034}\]
Answer: \(\tfrac{5}{34}\)

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Exercise 13.3, Question 12

A card is lost from a pack of 52. Two cards drawn from the rest are both diamonds. Find the probability the lost card is a diamond.
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  1. Lost diamond (\(\tfrac14\)): \(P = \dfrac{^{12}C_2}{^{51}C_2}\). Lost non-diamond (\(\tfrac34\)): \(P = \dfrac{^{13}C_2}{^{51}C_2}\).
  2. \[\dfrac{\frac14 \cdot 66}{\frac14 \cdot 66 + \frac34 \cdot 78} = \dfrac{66}{66 + 234}\]
Answer: \(\tfrac{11}{50}\)

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Exercise 13.3, Question 13

A speaks the truth with probability \(\tfrac45\). A coin is tossed and A reports a head. The probability there really was a head is: (A) \(\tfrac45\) (B) \(\tfrac12\) (C) \(\tfrac15\) (D) \(\tfrac25\)
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  1. \[\dfrac{\frac12 \cdot \frac45}{\frac12 \cdot \frac45 + \frac12 \cdot \frac15}\]
Answer: (A) \(\tfrac45\)

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Exercise 13.3, Question 14

\(A \subset B\) and \(P(B) \ne 0\). Which is correct? (A) \(P(A|B) = \dfrac{P(B)}{P(A)}\) (B) \(P(A|B) < P(A)\) (C) \(P(A|B) \ge P(A)\) (D) none
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  1. \(A \cap B = A\), so \(P(A|B) = \dfrac{P(A)}{P(B)} \ge P(A)\) because \(P(B) \le 1\).
Answer: (C)

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