Exercise 13.3 answers: Probability

Class 12 Maths14 questions

Exercise 13.3

14 questions · page 431 of the book

Question 1

“An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn.” · p. 431

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  1. Split by what the first ball was: red (chance 1/2) or black (chance 1/2).
  2. If the first was red, the urn now has 12 balls with 7 red, so the second is red with chance 7/12.
  3. If the first was black, the urn now has 12 balls with 5 red, so the second is red with chance 5/12.
  4. Total: (1/2)(7/12) + (1/2)(5/12) = 1/2.

Answer1/2

Watch this explained “An urn that changes between draws”, 12:50 into Partitioning the sample space, and totalling a probability across the parts

Question 2

“A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls.” · p. 431

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  1. Each bag is equally likely to be chosen: 1/2 each.
  2. Bag 1 gives red with chance 4/8 = 1/2; bag 2 gives red with chance 2/8 = 1/4.
  3. Chance of red overall = (1/2)(1/2) + (1/2)(1/4) = 3/8.
  4. By Bayes' theorem, P(bag 1 | red) = (1/2 × 1/2) ÷ 3/8 = 2/3.

Answer2/3

Watch this explained “Two bags, worked end to end”, 5:50 into Working back from an observed effect to the likelihood of each cause

Question 3

“Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel).” · p. 431

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  1. Two routes: hostel (chance 0.6) or day scholar (chance 0.4).
  2. A-grade chance along each route: 0.3 for hostel, 0.2 for day scholars.
  3. Chance of A grade overall = 0.6×0.3 + 0.4×0.2 = 0.18 + 0.08 = 0.26.
  4. By Bayes' theorem, P(hostel | A grade) = 0.18 ÷ 0.26 = 9/13.

Answer9/13

Watch this explained “Two bags, worked end to end”, 5:50 into Working back from an observed effect to the likelihood of each cause

Question 4

“In answering a question on a multiple choice test, a student either knows the answer or guesses.” · p. 431

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  1. Two routes: knows the answer (chance 3/4) or guesses (chance 1/4).
  2. If he knows, he is certainly correct; if he guesses, he is correct with chance 1/4.
  3. Chance of being correct overall = 3/4×1 + 1/4×1/4 = 3/4 + 1/16 = 13/16.
  4. By Bayes' theorem, P(knows | correct) = (3/4) ÷ (13/16) = 12/13.

Answer12/13

Watch this explained “Honest three times in four”, 14:15 into Working back from an observed effect to the likelihood of each cause

Question 5

“A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present.” · p. 431

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  1. Two routes: has the disease (chance 0.001) or is healthy (chance 0.999).
  2. The test is positive with chance 0.99 along the disease route, and 0.005 along the healthy route.
  3. Chance of a positive test overall = 0.001×0.99 + 0.999×0.005 = 0.00099 + 0.004995 = 0.005985.
  4. By Bayes' theorem, P(disease | positive) = 0.00099 ÷ 0.005985 = 22/133.

Answer22/133 (about 16.5%)

Watch this explained “A rare condition and a good test”, 8:33 into Working back from an observed effect to the likelihood of each cause

Question 6

“There are three coins. One is a two headed coin … a biased coin that comes up heads 75% of the time” · p. 432

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  1. Each coin is equally likely to be picked: 1/3 each.
  2. Chance of heads along each route: 1 (two-headed), 3/4 (biased), 1/2 (fair).
  3. Chance of heads overall = (1/3)(1 + 3/4 + 1/2) = (1/3)(9/4) = 3/4.
  4. By Bayes' theorem, P(two-headed | heads) = (1/3×1) ÷ (3/4) = 4/9.

Answer4/9

Watch this explained “Three boxes and a gold coin”, 7:04 into Working back from an observed effect to the likelihood of each cause

Question 7

“An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers.” · p. 432

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  1. Out of 12000 drivers, the chance of each type is 2000/12000, 4000/12000, 6000/12000 — that is 1/6, 1/3, 1/2.
  2. Accident chance along each route: 0.01 (scooter), 0.03 (car), 0.15 (truck).
  3. Chance of an accident overall = (1/6)(0.01) + (1/3)(0.03) + (1/2)(0.15) = 13/150.
  4. By Bayes' theorem, P(scooter | accident) = ((1/6)(0.01)) ÷ (13/150) = 1/52.

Answer1/52

Watch this explained “Three machines, one defective bolt”, 11:44 into Working back from an observed effect to the likelihood of each cause

Question 8

“machine A produced 60% of the items of output and machine B produced 40% of the items” · p. 432

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  1. Machine A makes 60% of items, Machine B makes 40%.
  2. Of A's items 2% are defective; of B's items 1% are defective.
  3. Chance an item is (from A and defective) = 60% × 2% = 1.2%.
  4. Chance an item is (from B and defective) = 40% × 1% = 0.4%.
  5. Total chance an item is defective = 1.2% + 0.4% = 1.6%.
  6. Chance it was made by B, given it is defective = 0.4% ÷ 1.6% = 1/4.

Answer1/4

Watch this explained “Three machines, one defective bolt”, 11:44 into Working back from an observed effect to the likelihood of each cause

Question 9

“probabilities that the first and the second groups will win are 0.6 and 0.4 respectively” · p. 432

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  1. First group wins with chance 0.6, second group wins with chance 0.4.
  2. If the first group wins, the chance of a new product is 0.7; if the second wins, it is 0.3.
  3. Chance of (first group wins and new product) = 0.6 × 0.7 = 0.42.
  4. Chance of (second group wins and new product) = 0.4 × 0.3 = 0.12.
  5. Total chance a new product is introduced = 0.42 + 0.12 = 0.54.
  6. Chance it was the second group, given a new product was introduced = 0.12 ÷ 0.54 = 2/9.

Answer2/9

Watch this explained “Three machines, one defective bolt”, 11:44 into Working back from an observed effect to the likelihood of each cause

Question 10

“If she gets a 5 or 6, she tosses a coin three times and notes the number of heads” · p. 432

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  1. Chance she rolls 1, 2, 3 or 4 = 4/6 = 2/3. Chance she rolls 5 or 6 = 2/6 = 1/3.
  2. If she rolled 1–4, she tosses one coin; chance of exactly one head = 1/2.
  3. If she rolled 5 or 6, she tosses three coins; chance of exactly one head out of three = 3 × (1/2)³ = 3/8.
  4. Chance of (rolled 1–4 and exactly one head) = 2/3 × 1/2 = 1/3.
  5. Chance of (rolled 5 or 6 and exactly one head) = 1/3 × 3/8 = 1/8.
  6. Total chance of exactly one head = 1/3 + 1/8 = 11/24.
  7. Chance she rolled 1, 2, 3 or 4, given exactly one head = (1/3) ÷ (11/24) = 8/11.

Answer8/11

Watch this explained “Two bags, worked end to end”, 5:50 into Working back from an observed effect to the likelihood of each cause

Question 11

“A is on the job for 50% of the time, B is on the job for 30% of the time” · p. 432

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  1. A works 50% of the time, B works 30%, C works 20%.
  2. A's defect rate is 1%, B's is 5%, C's is 7%.
  3. Chance of (defective and made by A) = 0.5 × 0.01 = 0.005.
  4. Chance of (defective and made by B) = 0.3 × 0.05 = 0.015.
  5. Chance of (defective and made by C) = 0.2 × 0.07 = 0.014.
  6. Total chance an item is defective = 0.005 + 0.015 + 0.014 = 0.034.
  7. Chance it was made by A, given it is defective = 0.005 ÷ 0.034 = 5/34.

Answer5/34

Watch this explained “Three machines, one defective bolt”, 11:44 into Working back from an observed effect to the likelihood of each cause

Question 12

“two cards are drawn and are found to be both diamonds” · p. 432

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  1. Before the card is lost, 13 of the 52 cards are diamonds, so the lost card is a diamond with chance 13/52 = 1/4, and not a diamond with chance 39/52 = 3/4.
  2. If the lost card was a diamond, 51 cards remain with 12 diamonds; the chance of drawing 2 diamonds from these is C(12,2)/C(51,2) = 66/1275.
  3. If the lost card was not a diamond, 51 cards remain with 13 diamonds; the chance of drawing 2 diamonds is C(13,2)/C(51,2) = 78/1275.
  4. Chance of (lost card diamond and drawing 2 diamonds) = 1/4 × 66/1275.
  5. Chance of (lost card not diamond and drawing 2 diamonds) = 3/4 × 78/1275.
  6. Total chance of drawing 2 diamonds = (1/4 × 66 + 3/4 × 78) / 1275 = 75/1275.
  7. Chance the lost card was a diamond, given 2 diamonds were drawn = (1/4 × 66/1275) ÷ (75/1275) = 11/50.

Answer11/50

Watch this explained “Two bags, worked end to end”, 5:50 into Working back from an observed effect to the likelihood of each cause

Question 13

“Probability that A speaks truth is 4/5. A coin is tossed. A reports that a head appears.” · p. 432

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  1. A speaks the truth with chance 4/5, so A lies with chance 1/5.
  2. The coin is fair, so it actually shows heads with chance 1/2 and tails with chance 1/2.
  3. A reports 'head' either by telling the truth after an actual head, or by lying after an actual tail.
  4. Chance of (actual head and A reports head) = 1/2 × 4/5 = 4/10.
  5. Chance of (actual tail and A reports head) = 1/2 × 1/5 = 1/10.
  6. Total chance A reports head = 4/10 + 1/10 = 5/10.
  7. Chance it was actually a head, given A reports head = (4/10) ÷ (5/10) = 4/5.

Answer4/5 — option (A)

Watch this explained “Honest three times in four”, 14:15 into Working back from an observed effect to the likelihood of each cause

Question 14

“A and B are two events such that A ⊂ B and P(B) ≠ 0” · p. 433

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  1. Since A ⊂ B, every outcome in A is also in B, so A ∩ B = A.
  2. By definition, P(A|B) = P(A ∩ B) / P(B) = P(A) / P(B).
  3. Because P(B) is at most 1, dividing P(A) by P(B) can only keep it the same or make it bigger.
  4. So P(A|B) = P(A)/P(B) ≥ P(A).

AnswerOption (C): P(A|B) ≥ P(A)

Watch this explained “The definition”, 5:06 into Being told one event happened shrinks the sample space you count against

Every question here was solved twice, separately, by two different AI models, and each answer was put back into the question by a computer program to check it. Where the two disagreed, a stronger model solved it again and the computer check had to pass on its answer. A question about reasoning rather than a number is shown as “one way to think about it”, and anything not yet proven says so instead of guessing. Each answer links to the moment in the video that teaches it.

We quote only enough of each question to find it: keep your NCERT book open, or open this chapter in NCERT’s PDF. Spotted a mistake? Tell us.

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