Miscellaneous Exercise answers: Probability

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(i) A is a subset of B

1. A is a subset of B, so every outcome of A already lies in B: A ∩ B = A.
2. P(B|A) = P(A ∩ B) / P(A) = P(A) / P(A) = 1.

Answer1

(ii) A ∩ B = φ

1. A ∩ B = φ means A and B share no outcome, so P(A ∩ B) = 0.
2. P(B|A) = P(A ∩ B) / P(A) = 0 / P(A) = 0.

Answer0

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

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(i) both children are males, if it is known that at least one

1. List the four equally likely outcomes for two children in birth order: (boy,boy), (boy,girl), (girl,boy), (girl,girl).
2. 'At least one is male' rules out (girl,girl), leaving 3 equally likely outcomes.
3. Only 1 of these 3 outcomes, (boy,boy), has both children male.
4. Chance both are male, given at least one is male = 1/3.

Answer1/3

(ii) both children are females, if it is known that the elder

1. Using the same four outcomes, 'the elder child is female' keeps (girl,boy) and (girl,girl) — 2 outcomes.
2. Only 1 of these 2 outcomes, (girl,girl), has both children female.
3. Chance both are female, given the elder child is female = 1/2.

Answer1/2

Watch this explained “Five worked shapes”, 8:09 into Being told one event happened shrinks the sample space you count against

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1. Assume equal numbers of men and women, so each is chosen with chance 1/2.
2. 5% of men have grey hair; 0.25% of women have grey hair.
3. Chance of (male and grey hair) = 1/2 × 5/100 = 1/40.
4. Chance of (female and grey hair) = 1/2 × 0.25/100 = 1/800.
5. Total chance of grey hair = 1/40 + 1/800 = 21/800.
6. Chance the person is male, given grey hair = (1/40) ÷ (21/800) = 20/21.

Answer20/21

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

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1. Each person is right-handed with chance 0.9 and left-handed with chance 0.1, independently. Let X be the number of right-handed people among the 10.
2. The chance that exactly k of the 10 are right-handed is C(10,k) × (0.9)^k × (0.1)^(10−k).
3. 'At most 6' is everything except 7, 8, 9 or 10, so P(X ≤ 6) = 1 − [P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)].
4. P(X = 10) = (0.9)^10 = 0.3486784401.
5. P(X = 9) = 10 × (0.9)^9 × 0.1 = 0.387420489.
6. P(X = 8) = 45 × (0.9)^8 × (0.1)^2 = 0.1937102445.
7. P(X = 7) = 120 × (0.9)^7 × (0.1)^3 = 0.057395628.
8. These four add to 0.9872048016.
9. P(X ≤ 6) = 1 − 0.9872048016 = 0.0127951984 = 7996999/625000000.

Answer7996999/625000000 = 0.0127951984 (about 0.0128)

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1. A leap year has 366 days, which is 52 complete weeks (364 days) plus 2 extra days.
2. These 2 extra days can be any one of 7 equally likely consecutive pairs: (Sun,Mon), (Mon,Tue), (Tue,Wed), (Wed,Thu), (Thu,Fri), (Fri,Sat), (Sat,Sun).
3. A 53rd Tuesday happens only if Tuesday is one of these 2 extra days, which happens in 2 of the 7 pairs.
4. Chance of 53 Tuesdays = 2/7.

Answer2/7

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1. Each box is equally likely to be picked: chance 1/4 each.
2. Box A has 1 red ball out of 10, so chance of (box A and red) = 1/4 × 1/10 = 1/40.
3. Box B has 6 red out of 10, so chance of (box B and red) = 1/4 × 6/10 = 6/40.
4. Box C has 8 red out of 10, so chance of (box C and red) = 1/4 × 8/10 = 8/40.
5. Box D has 0 red out of 10, so it contributes 0.
6. Total chance of drawing red = 1/40 + 6/40 + 8/40 + 0 = 15/40 = 3/8.
7. Chance the marble came from box A, given it is red = (1/40) ÷ (3/8) = 1/15.
8. Chance the marble came from box B, given it is red = (6/40) ÷ (3/8) = 2/5.
9. Chance the marble came from box C, given it is red = (8/40) ÷ (3/8) = 8/15.

AnswerBox A: 1/15, Box B: 2/5, Box C: 8/15

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

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1. Chance of choosing meditation and yoga = 1/2; chance of choosing the drug = 1/2.
2. Base chance of a heart attack is 40%. Meditation and yoga cut this risk by 30%, so the chance of a heart attack after yoga = 40% × (1 − 30%) = 28%.
3. The drug cuts the risk by 25%, so the chance of a heart attack after the drug = 40% × (1 − 25%) = 30%.
4. Chance of (chose yoga and heart attack) = 1/2 × 28% = 14%.
5. Chance of (chose drug and heart attack) = 1/2 × 30% = 15%.
6. Total chance of a heart attack = 14% + 15% = 29%.
7. Chance the patient followed meditation and yoga, given a heart attack occurred = 14/29.

Answer14/29

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

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1. The determinant of a 2×2 matrix with entries a, b (top row) and c, d (bottom row) is a×d − b×c.
2. Each entry is 0 or 1 independently with chance 1/2, so a×d − b×c can only be −1, 0 or 1.
3. It is positive only when a×d − b×c = 1, which needs a×d = 1 (so a = d = 1) and b×c = 0.
4. Chance a = d = 1 is 1/2 × 1/2 = 1/4.
5. Chance b×c = 0 is 1 minus the chance b = c = 1, which is 1 − 1/4 = 3/4.
6. Since all four entries are chosen independently, chance the determinant is positive = 1/4 × 3/4 = 3/16.

Answer3/16

Watch this explained “At least one of the two”, 13:08 into Independence, and why it is a different idea from having no outcome in common

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(i) P(A fails|B has failed)

1. P(B fails) is B failing alone plus B failing together with A: 0.15 + 0.15 = 0.30.
2. P(A fails | B has failed) = P(A and B fail) ÷ P(B fails) = 0.15 ÷ 0.30 = 1/2.

Answer1/2

(ii) P(A fails alone)

1. 'A fails alone' means A fails but B does not, which is P(A fails) minus P(A and B fail).
2. P(A fails alone) = 0.2 − 0.15 = 0.05 = 1/20.

Answer1/20

Watch this explained “Four pieces, read off two events”, 6:05 into Partitioning the sample space, and totalling a probability across the parts

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1. Bag I has 3 red and 4 black balls (7 total), so the transferred ball is red with chance 3/7 and black with chance 4/7.
2. If a red ball was transferred, Bag II becomes 5 red and 5 black (10 total), so the chance of then drawing red = 5/10 = 1/2.
3. If a black ball was transferred, Bag II becomes 4 red and 6 black (10 total), so the chance of then drawing red = 4/10 = 2/5.
4. Chance of (transfer red and draw red) = 3/7 × 1/2 = 3/14.
5. Chance of (transfer black and draw red) = 4/7 × 2/5 = 8/35.
6. Total chance of drawing red = 3/14 + 8/35 = 31/70.
7. Chance the transferred ball was black, given a red ball was drawn = (8/35) ÷ (31/70) = 16/31.

Answer16/31

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

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1. By definition, P(B|A) = P(A ∩ B) / P(A). Since this equals 1, P(A ∩ B) = P(A).
2. A ∩ B is always a part of A, so if P(A ∩ B) equals the whole of P(A), every outcome of A must also lie in B.
3. That means A ⊂ B.

AnswerOption (A): A ⊂ B

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

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1. P(A|B) > P(A) means P(A ∩ B)/P(B) > P(A), so P(A ∩ B) > P(A) × P(B).
2. Divide both sides by P(A): P(A ∩ B)/P(A) > P(B).
3. The left side is exactly P(B|A), so P(B|A) > P(B).

AnswerOption (C): P(B|A) > P(B)

Watch this explained “The other way round”, 1:51 into Multiplying along a chain of dependent draws, and extending it past two events

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1. P(A) + P(B) − P(A and B) is the chance that A or B (or both) happens, P(A ∪ B).
2. The question says this equals P(A), so P(A ∪ B) = P(A), meaning B adds nothing new beyond A.
3. Rearranging the given equation: P(B) − P(A and B) = 0, so P(A and B) = P(B).
4. P(A|B) = P(A and B) / P(B) = P(B)/P(B) = 1.

AnswerOption (B): P(A|B) = 1

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