Exercise 13.1 answers: Probability

Open NCERT p. 413Matches NCERT’s answer

1. P(E|F) is P(E∩F) divided by P(F).
2. P(E|F) = 0.2 ÷ 0.3 = 2/3.
3. P(F|E) is P(E∩F) divided by P(E).
4. P(F|E) = 0.2 ÷ 0.6 = 1/3.

AnswerP(E|F) = 2/3, P(F|E) = 1/3

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Open NCERT p. 413Matches NCERT’s answer

1. P(A|B) = P(A∩B) ÷ P(B).
2. P(A|B) = 0.32 ÷ 0.5 = 0.64 = 16/25.

AnswerP(A|B) = 16/25

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Open NCERT p. 413Matches NCERT’s answer

(i)

1. P(A∩B) = P(A) × P(B|A).
2. P(A∩B) = 0.8 × 0.4 = 0.32 = 8/25.

Answer8/25

(ii)

1. P(A|B) = P(A∩B) ÷ P(B).
2. P(A|B) = 0.32 ÷ 0.5 = 0.64 = 16/25.

Answer16/25

(iii)

1. P(A∪B) = P(A) + P(B) − P(A∩B).
2. P(A∪B) = 0.8 + 0.5 − 0.32 = 0.98 = 49/50.

Answer49/50

Watch this explained “Clear the denominator”, 0:48 into Multiplying along a chain of dependent draws, and extending it past two events

Open NCERT p. 413Matches NCERT’s answer

1. 2P(A) = 5/13, so P(A) = 5/26.
2. P(A∩B) = P(B) × P(A|B) = 5/13 × 2/5 = 2/13.
3. P(A∪B) = P(A) + P(B) − P(A∩B).
4. P(A∪B) = 5/26 + 10/26 − 4/26 = 11/26.

AnswerP(A∪B) = 11/26

Watch this explained “Clear the denominator”, 0:48 into Multiplying along a chain of dependent draws, and extending it past two events

Open NCERT p. 413Matches NCERT’s answer

(i)

1. P(A∩B) = P(A) + P(B) − P(A∪B).
2. P(A∩B) = 6/11 + 5/11 − 7/11 = 4/11.

Answer4/11

(ii)

1. P(A|B) = P(A∩B) ÷ P(B) = (4/11) ÷ (5/11).
2. P(A|B) = 4/5.

Answer4/5

(iii)

1. P(B|A) = P(A∩B) ÷ P(A) = (4/11) ÷ (6/11).
2. P(B|A) = 2/3.

Answer2/3

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

Open NCERT p. 413Matches NCERT’s answer

(i) head on third toss , F : heads on first two tosses

1. 8 equally likely outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.
2. F (heads on first two tosses) = {HHH, HHT}, so P(F) = 2/8.
3. E (head on third toss) ∩ F = {HHH}, so P(E∩F) = 1/8.
4. P(E|F) = (1/8) ÷ (2/8) = 1/2.

Answer1/2

(ii) at least two heads , F : at most two heads

1. F (at most two heads) = every outcome except HHH, so P(F) = 7/8.
2. E (at least two heads) ∩ F = {HHT, HTH, THH}, so P(E∩F) = 3/8.
3. P(E|F) = (3/8) ÷ (7/8) = 3/7.

Answer3/7

(iii) at most two tails , F : at least one tail

1. F (at least one tail) = every outcome except HHH, so P(F) = 7/8.
2. E (at most two tails) = every outcome except TTT, so E∩F excludes both HHH and TTT, giving 6 outcomes.
3. P(E|F) = (6/8) ÷ (7/8) = 6/7.

Answer6/7

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Open NCERT p. 414Matches NCERT’s answer

(i) tail appears on one coin, F : one coin shows head

1. Tossing two coins gives 4 equally likely outcomes: HH, HT, TH, TT.
2. 'Tail appears on one coin' means exactly one tail, so E = {HT, TH}. 'One coin shows head' means exactly one head, so F = {HT, TH}. (The book writes 'at least' when it means at least.)
3. E∩F = {HT, TH}, so P(E∩F) = 2/4, and P(F) = 2/4.
4. P(E|F) = (2/4) ÷ (2/4) = 1.

Answer1

(ii) E : no tail appears, F : no head appears

1. E (no tail appears) = {HH}. F (no head appears) = {TT}, so P(F) = 1/4.
2. E and F share no outcome, so P(E∩F) = 0.
3. P(E|F) = 0 ÷ (1/4) = 0.

Answer0

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1. F fixes the first throw as 6 and the second as 5; the third throw can be any of the 6 faces.
2. Among these 6 equally likely cases, only the one where the third throw is 4 is in E.
3. P(E|F) = 1/6.

AnswerP(E|F) = 1/6

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Open NCERT p. 414Matches NCERT’s answer

1. 6 equally likely arrangements of Mother, Father, Son in a line.
2. F (father in middle) leaves 2 arrangements: Son-Father-Mother and Mother-Father-Son.
3. In both of these, the son is on one end, so both are also in E.
4. P(E|F) = 2/2 = 1.

AnswerP(E|F) = 1

Watch this explained “A ratio of two counts”, 3:25 into Being told one event happened shrinks the sample space you count against

Open NCERT p. 414Matches NCERT’s answer

(a) the black die resulted in a 5

1. Black die shows 5, so the red die is any of 1 to 6 — 6 equally likely cases.
2. Sum > 9 means red > 4, so red = 5 or 6 — 2 cases.
3. Conditional probability = 2/6 = 1/3.

Answer1/3

(b) the red die resulted in a number less than 4

1. Red die < 4, so red ∈ {1, 2, 3} and black ∈ {1,...,6} — 18 equally likely cases.
2. Sum = 8 happens only for (black, red) = (6, 2) and (5, 3) — 2 cases.
3. Conditional probability = 2/18 = 1/9.

Answer1/9

Watch this explained “Which survivors count”, 2:33 into Being told one event happened shrinks the sample space you count against

Open NCERT p. 414Matches NCERT’s answer

(i)

1. E∩F = {3}, so P(E∩F) = 1/6. P(F) = 2/6, P(E) = 3/6.
2. P(E|F) = (1/6) ÷ (2/6) = 1/2.
3. P(F|E) = (1/6) ÷ (3/6) = 1/3.

AnswerP(E|F) = 1/2, P(F|E) = 1/3

(ii)

1. E∩G = {3,5}, so P(E∩G) = 2/6. P(G) = 4/6, P(E) = 3/6.
2. P(E|G) = (2/6) ÷ (4/6) = 1/2.
3. P(G|E) = (2/6) ÷ (3/6) = 2/3.

AnswerP(E|G) = 1/2, P(G|E) = 2/3

(iii)

1. E∪F = {1,2,3,5}. (E∪F)∩G = {2,3,5}, so P((E∪F)∩G) = 3/6.
2. P((E∪F)|G) = (3/6) ÷ (4/6) = 3/4.
3. E∩F = {3}. (E∩F)∩G = {3}, so P((E∩F)∩G) = 1/6.
4. P((E∩F)|G) = (1/6) ÷ (4/6) = 1/4.

AnswerP((E∪F)|G) = 3/4, P((E∩F)|G) = 1/4

Watch this explained “All of it, on one die”, 11:38 into The rules conditional probability inherits, including addition on a restricted space

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(i) the youngest is a girl

1. 4 equally likely outcomes for (elder, younger): BB, BG, GB, GG.
2. Youngest is a girl: {BG, GG} — 2 outcomes.
3. Both girls among these: {GG} — 1 outcome.
4. Conditional probability = 1/2.

Answer1/2

(ii) at least one is a girl

1. At least one girl: {BG, GB, GG} — 3 outcomes.
2. Both girls among these: {GG} — 1 outcome.
3. Conditional probability = 1/3.

Answer1/3

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1. Total multiple choice questions = 500 (easy) + 400 (difficult) = 900.
2. Of these, 500 are easy.
3. Probability = 500/900 = 5/9.

Answer5/9

Watch this explained “A ratio of two counts”, 3:25 into Being told one event happened shrinks the sample space you count against

Open NCERT p. 414Matches NCERT’s answer

1. 36 equally likely outcomes; removing the 6 doubles leaves 30 outcomes with different numbers.
2. Sum = 4 with different numbers: (1,3) and (3,1) — 2 outcomes. ((2,2) is a double, so it is excluded.)
3. Conditional probability = 2/30 = 1/15.

Answer1/15

Watch this explained “Told: the first is tails”, 1:48 into Being told one event happened shrinks the sample space you count against

Open NCERT p. 414Matches NCERT’s answer

1. A coin is tossed only when the first throw is 1, 2, 4 or 5.
2. 'At least one die shows a 3' can only happen when the first throw is 3 (then a second die is thrown), or the first throw is 6 and the second die shows 3.
3. So whenever a coin is tossed, the first throw is not 3, and no second die was thrown — 'at least one die shows a 3' is false.
4. The two events share no outcome, so the conditional probability is 0.

Answer0

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Open NCERT p. 414Checked by computer

1. P(A|B) = P(A∩B) ÷ P(B), and this needs P(B) ≠ 0.
2. Here P(B) = 0, so P(A|B) cannot be worked out — it is not defined.

Answer(C) not defined

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Open NCERT p. 415Matches NCERT’s answer

1. P(A|B) = P(A∩B) ÷ P(B) and P(B|A) = P(A∩B) ÷ P(A).
2. We are told these are equal, so P(A∩B) ÷ P(B) = P(A∩B) ÷ P(A).
3. When A and B can happen together, P(A∩B) is not 0, so divide both sides by it: 1 ÷ P(B) = 1 ÷ P(A), which gives P(A) = P(B).
4. The other options do not have to be true. On a die, take A = {1, 2} and B = {2, 3}: P(A|B) = 1/2 = P(B|A), yet A is not inside B, A ≠ B, and A∩B = {2} is not empty.

Answer(D) P(A) = P(B)

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