Exercise 14.2 answers: Probability

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(a)

1. Add the seven numbers: 0.1+0.01+0.05+0.03+0.01+0.2+0.6 = 1.
2. Every value is between 0 and 1, and the total is exactly 1.

AnswerValid — non-negative and adds to 1.

(b)

1. Each of the seven outcomes gets 1/7.
2. Seven sevenths add to exactly 1, and each value is between 0 and 1.

AnswerValid — an equal split of probability among all seven outcomes.

(c)

1. Add the seven values: 0.1+0.2+0.3+0.4+0.5+0.6+0.7 = 2.8.

AnswerNot valid — the probabilities add up to 2.8, not 1.

(d)

1. Two of the values, −0.1 and −0.2, are negative.

AnswerNot valid — a probability can never be negative.

(e)

1. Add the seven values: (1+2+3+4+5+6+15)/14 = 36/14, which is more than 1.
2. The last value, 15/14, is itself bigger than 1.

AnswerNot valid — the probabilities add to 36/14, which is more than 1.

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1. List all outcomes of tossing a coin twice: HH, HT, TH, TT, each equally likely.
2. ‘At least one tail’ fails only for HH, the one outcome with no tail at all.
3. So P(at least one tail) = 1 − P(HH) = 1 − 1/4.

Answer3/4

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(i) A prime number will appear

1. The prime faces on a die are 2, 3 and 5 — 3 outcomes out of 6.

Answer1/2

(ii) A number greater than or equal to 3

1. 3, 4, 5 and 6 qualify — 4 outcomes out of 6.

Answer2/3

(iii) A number less than or equal to one

1. Only 1 qualifies — 1 outcome out of 6.

Answer1/6

(iv) A number more than 6 will appear

1. No face on a die is more than 6.

Answer0

(v) A number less than 6 will appear

1. 1, 2, 3, 4 and 5 qualify — 5 outcomes out of 6.

Answer5/6

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(a) How many points are there in the sample space?

1. A pack has 52 different cards, each equally likely to be drawn.

Answer52

(b) the card is an ace of spades

1. Only one card is exactly the ace of spades, out of 52.

Answer1/52

(c)(i) an ace

1. There are 4 aces in the pack, one per suit.

Answer4/52 = 1/13

(c)(ii) black card

1. Half the pack is black — 26 clubs and spades.

Answer26/52 = 1/2

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(i) 3

1. The coin shows 1 or 6, and the die shows 1 to 6, giving 12 equally likely pairs.
2. A sum of 3 only happens as (coin = 1, die = 2).

Answer1/12

(ii) 12

1. A sum of 12 only happens as (coin = 6, die = 6).

Answer1/12

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1. The council has 4 + 6 = 10 members in total.
2. 6 of these 10 members are women.

Answer6/10 = 3/5

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1. There are 2⁴ = 16 equally likely sequences of 4 tosses.
2. With h heads (and 4−h tails), the money is h × ₹1 − (4−h) × ₹1.50.
3. Working this out for h = 0, 1, 2, 3, 4 gives −₹6, −₹3.50, −₹1, ₹1.50 and ₹4 — five different amounts.
4. The number of sequences giving h heads is ⁴Cₕ: 1, 4, 6, 4, 1 out of 16.
5. So, in the same order as the amounts, the probabilities are 1/16, 1/4, 3/8, 1/4 and 1/16.

Answer5 different amounts: −₹6, −₹3.50, −₹1, ₹1.50, ₹4, with probabilities 1/16, 1/4, 3/8, 1/4, 1/16.

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(i) 3 heads

1. Only HHH gives 3 heads — 1 outcome out of 8.

Answer1/8

(ii) 2 heads

1. HHT, HTH and THH give exactly 2 heads — 3 outcomes.

Answer3/8

(iii) atleast 2 heads

1. Add the 3 outcomes with 2 heads to the 1 outcome with 3 heads — 4 outcomes.

Answer1/2

(iv) atmost 2 heads

1. Every outcome except HHH has at most 2 heads — 7 outcomes.

Answer7/8

(v) no head

1. Only TTT has no head.

Answer1/8

(vi) 3 tails

1. Only TTT has 3 tails — the same single outcome as ‘no head’.

Answer1/8

(vii) exactly two tails

1. Exactly 2 tails means exactly 1 head — HTT, THT, TTH — 3 outcomes.

Answer3/8

(viii) no tail

1. Only HHH has no tail.

Answer1/8

(ix) atmost two tails

1. Every outcome except TTT has at most 2 tails — 7 outcomes.

Answer7/8

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1. For any event, P(A) + P(not A) = 1.
2. So P(not A) = 1 − 2/11.

Answer9/11

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(i) a vowel

1. The word ASSASSINATION has 13 letters in all.
2. The vowels among them are A (×3), I (×2) and O (×1) — 6 vowels out of 13.

Answer6/13

(ii) a consonant

1. The consonants are S (×4), N (×2) and T (×1) — 7 out of 13.

Answer7/13

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1. The person chooses 6 different numbers from the numbers 1 to 20.
2. Since order does not matter, the total number of ways to choose 6 numbers from 20 is C(20,6).
3. C(20,6) = 20!/(6! × 14!) = (20 × 19 × 18 × 17 × 16 × 15)/(6 × 5 × 4 × 3 × 2 × 1)
4. C(20,6) = 27907200/720 = 38760
5. There is only 1 way to choose the exact 6 numbers fixed by the committee.
6. P(winning) = 1/38760

Answer1/38760

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(i) P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6

1. A ∩ B is part of both A and B, so P(A ∩ B) can never be more than the smaller of P(A) and P(B).
2. Here P(A ∩ B) = 0.6 is bigger than P(A) = 0.5 — that is impossible.

AnswerNot consistently defined.

(ii) P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8

1. Using P(A ∪ B) = P(A) + P(B) − P(A ∩ B), the implied overlap is 0.5 + 0.4 − 0.8 = 0.1.
2. 0.1 lies between 0 and the smaller of 0.5 and 0.4, so this is possible.

AnswerConsistently defined.

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(i)

1. Use P(A∪B) = P(A) + P(B) − P(A∩B).
2. P(A) = 1/3, P(B) = 1/5, P(A∩B) = 1/15.
3. P(A∪B) = 1/3 + 1/5 − 1/15 = 5/15 + 3/15 − 1/15 = 7/15.

Answer7/15

(ii)

1. Rearrange the same rule: P(B) = P(A∪B) − P(A) + P(A∩B).
2. P(A∪B) = 0.6, P(A) = 0.35, P(A∩B) = 0.25.
3. P(B) = 0.6 − 0.35 + 0.25 = 0.5.

Answer1/2

(iii)

1. Rearrange the rule: P(A∩B) = P(A) + P(B) − P(A∪B).
2. P(A) = 0.5, P(B) = 0.35, P(A∪B) = 0.7.
3. P(A∩B) = 0.5 + 0.35 − 0.7 = 0.15.

Answer3/20

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1. A and B are mutually exclusive, so they share no outcome.
2. For mutually exclusive events, P(A or B) = P(A) + P(B).
3. P(A or B) = 3/5 + 1/5 = 4/5.

Answer4/5

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(i) P(E or F)

1. Use the addition rule: P(E or F) = P(E) + P(F) − P(E and F).
2. P(E or F) = 1/4 + 1/2 − 1/8 = 2/8 + 4/8 − 1/8 = 5/8.

Answer5/8

(ii) P(not E and not F)

1. "Not E and not F" is the complement of "E or F".
2. P(not E and not F) = 1 − P(E or F) = 1 − 5/8 = 3/8.

Answer3/8

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1. "Not E or not F" is the complement of "E and F" (De Morgan's law).
2. So P(E and F) = 1 − P(not E or not F) = 1 − 0.25 = 0.75.
3. E and F are mutually exclusive only if P(E and F) = 0.
4. Since P(E and F) = 0.75, which is not 0, E and F are not mutually exclusive.

AnswerNo

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(i) P(not A)

1. P(not A) = 1 − P(A) = 1 − 0.42 = 0.58.

Answer29/50

(ii) P(not B)

1. P(not B) = 1 − P(B) = 1 − 0.48 = 0.52.

Answer13/25

(iii) P(A or B)

1. Use the addition rule: P(A or B) = P(A) + P(B) − P(A and B).
2. P(A or B) = 0.42 + 0.48 − 0.16 = 0.74.

Answer37/50

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1. Let M = studies Mathematics, B = studies Biology.
2. P(M) = 40/100, P(B) = 30/100, P(M and B) = 10/100.
3. Use the addition rule: P(M or B) = P(M) + P(B) − P(M and B).
4. P(M or B) = 0.4 + 0.3 − 0.1 = 0.6 = 3/5.

Answer3/5

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1. Let A = passes first exam, B = passes second exam.
2. P(A) = 0.8, P(B) = 0.7, P(A or B) = 0.95.
3. Use the addition rule: P(A and B) = P(A) + P(B) − P(A or B).
4. P(A and B) = 0.8 + 0.7 − 0.95 = 0.55 = 11/20.

Answer11/20

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1. Let E = passes English, H = passes Hindi.
2. P(E and H) = 0.5, P(passes neither) = 0.1, so P(E or H) = 1 − 0.1 = 0.9.
3. P(E) = 0.75.
4. Use the addition rule: P(H) = P(E or H) − P(E) + P(E and H).
5. P(H) = 0.9 − 0.75 + 0.5 = 0.65 = 13/20.

Answer13/20

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(i) The student opted for NCC or NSS

1. P(NCC) = 30/60, P(NSS) = 32/60, P(both) = 24/60.
2. P(NCC or NSS) = 30/60 + 32/60 − 24/60 = 38/60 = 19/30.

Answer19/30

(ii) opted neither NCC nor NSS

1. "Neither" is the complement of "NCC or NSS".
2. P(neither) = 1 − 19/30 = 11/30.

Answer11/30

(iii) opted NSS but not NCC

1. "NSS but not NCC" removes the students who did both from the NSS group.
2. P(NSS but not NCC) = P(NSS) − P(both) = 32/60 − 24/60 = 8/60 = 2/15.

Answer2/15

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