Exercise 14.1 answers: Probability

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1. E holds only the outcome 4, so E = {4}.
2. F holds the even faces 2, 4 and 6, so F = {2, 4, 6}.
3. Check what E and F have in common: the outcome 4 is in both.
4. Since they share an outcome, both events can happen on the same roll.

AnswerNo, E and F are not mutually exclusive — they share the outcome 4.

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(i) a number less than 7

1. A die shows a number from 1 to 6.
2. Every one of these six numbers is less than 7.

AnswerA = {1, 2, 3, 4, 5, 6}

(ii) a number greater than 7

1. A die can only show 1 to 6.
2. None of these is greater than 7.

AnswerB = { } (the empty set)

(iii) a multiple of 3

1. Check each face from 1 to 6 for being a multiple of 3.
2. Only 3 and 6 qualify.

AnswerC = {3, 6}

(iv) a number less than 4

1. Check each face from 1 to 6.
2. 1, 2 and 3 are less than 4.

AnswerD = {1, 2, 3}

(v) an even number greater than 4

1. The even faces are 2, 4 and 6.
2. Of these, only 6 is greater than 4.

AnswerE = {6}

(vi) a number not less than 3

1. ‘Not less than 3’ means 3 or more.
2. That gives 3, 4, 5 and 6.

AnswerF = {3, 4, 5, 6}

Also

1. A ∪ B combines every outcome in A or B: since B is empty, A ∪ B = A.
2. A ∩ B keeps outcomes in both: since B is empty, A ∩ B = { }.
3. B ∪ C combines B and C: since B is empty, B ∪ C = C = {3, 6}.
4. E ∩ F keeps outcomes in both E = {6} and F = {3, 4, 5, 6}: only 6 qualifies.
5. D ∩ E keeps outcomes in both D = {1, 2, 3} and E = {6}: nothing matches.
6. A − C removes C's outcomes from A: {1,2,3,4,5,6} minus {3,6} leaves {1,2,4,5}.
7. D − E removes E's outcomes from D: {1,2,3} minus {6} leaves {1,2,3} unchanged.
8. F′ is everything in {1,...,6} that is not in F: that leaves {1, 2}.
9. E ∩ F′ keeps outcomes in both E = {6} and F′ = {1, 2}: nothing matches.

AnswerA∪B = {1,2,3,4,5,6}; A∩B = { }; B∪C = {3,6}; E∩F = {6}; D∩E = { }; A−C = {1,2,4,5}; D−E = {1,2,3}; F′ = {1,2}; E∩F′ = { }

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1. Build the sample space: all 36 ordered pairs (x, y) with x, y from 1 to 6.
2. A is every pair whose sum is more than 8 — this gives 10 pairs.
3. B is every pair where a 2 shows on either die — this gives 11 pairs.
4. C is every pair whose sum is at least 7 and a multiple of 3, i.e. sum 9 or 12 — this gives 5 pairs.
5. A ∩ B: no pair with a 2 can reach a sum above 8 (the biggest such sum is 2+6=8), so A and B share nothing.
6. B ∩ C: the sums in C are 9 and 12, and no pair reaching those sums has a 2 in it, so B and C share nothing.
7. A ∩ C: every pair in C also has sum above 8, so A and C share all of C — not empty.

AnswerA and B are mutually exclusive, and B and C are mutually exclusive; A and C are not.

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(i) mutually exclusive?

1. List the outcomes: A = {HHH}, B = {HHT, HTH, THH}, C = {TTT}, D = {HHH, HHT, HTH, HTT}.
2. Check each pair for a common outcome: A∩B, A∩C, B∩C and C∩D all come out empty.
3. A∩D shares HHH, and B∩D shares HHT and HTH, so these two pairs are not mutually exclusive.

AnswerA&B, A&C, B&C and C&D are mutually exclusive; A&D and B&D are not.

(ii) simple?

1. A simple event holds exactly one outcome.
2. A = {HHH} and C = {TTT} each hold one outcome, so both are simple.
3. B holds 3 outcomes and D holds 4, so neither is simple.

AnswerA and C are simple events; B and D are not.

(iii) Compound?

1. A compound event holds more than one outcome.
2. B holds 3 outcomes and D holds 4, so both are compound.
3. A and C hold only one outcome each, so neither is compound.

AnswerB and D are compound events; A and C are not.

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(i) Two events which are mutually exclusive

1. Many different answers are correct for every part of this question: any events that meet the stated conditions will do. One example is shown for each part.
2. The 8 outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.
3. Take A: ‘all three coins show heads’ = {HHH}.
4. Take B: ‘all three coins show tails’ = {TTT}.
5. HHH and TTT cannot both happen on the same toss, so A ∩ B = ∅.

AnswerOne example: A = {HHH} and B = {TTT} are mutually exclusive.

(ii) mutually exclusive and exhaustive

1. Sort the eight outcomes by how many heads they hold: none, exactly one, or at least two.
2. A: no head = {TTT}. B: exactly one head = {HTT, THT, TTH}. C: at least two heads = {HHT, HTH, THH, HHH}.
3. No outcome fits two of these groups, and together they use up all 8 outcomes (1 + 3 + 4 = 8).

AnswerOne example: A = {TTT}, B = {HTT, THT, TTH}, C = {HHT, HTH, THH, HHH} — mutually exclusive and exhaustive.

(iii) which are not mutually exclusive

1. Take A: ‘at most two heads’ = every outcome except HHH.
2. Take B: ‘at least two heads’ = {HHT, HTH, THH, HHH}.
3. HHT, HTH and THH lie in both A and B, so A and B can happen together.

AnswerOne example: A = at most two heads, B = at least two heads — not mutually exclusive.

(iv) mutually exclusive but not exhaustive

1. Take A: no head = {TTT}.
2. Take B: exactly one head = {HTT, THT, TTH}.
3. These share no outcome, but together they leave out the outcomes with two or three heads.

AnswerOne example: A = {TTT}, B = {HTT, THT, TTH} — mutually exclusive but not exhaustive.

(v) mutually exclusive but not exhaustive

1. Take A: no head = {TTT}, B: exactly one head = {HTT, THT, TTH}, C: exactly two heads = {HHT, HTH, THH}.
2. No two of these share an outcome.
3. Together they leave out HHH (three heads), so they are not exhaustive.

AnswerOne example: A = {TTT}, B = {HTT, THT, TTH}, C = {HHT, HTH, THH} — mutually exclusive but not exhaustive.

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(i) A′

1. A′ is everything outside A.
2. Since A is ‘first die even’, A′ is ‘first die odd’.

AnswerA′ = {(x, y) : x is odd} — 18 outcomes, the same as B.

(ii) not B

1. ‘not B’ is B′, everything outside B.
2. Since B is ‘first die odd’, B′ is ‘first die even’.

Answernot B = {(x, y) : x is even} — 18 outcomes, the same as A.

(iii) A or B

1. A or B is A ∪ B.
2. Every roll has the first die either even or odd, so every outcome lands in A or B.

AnswerA or B = the whole sample space, all 36 outcomes.

(iv) A and B

1. A and B is A ∩ B.
2. No roll has the first die both even and odd at once.

AnswerA and B = { } (the empty set).

(v) A but not C

1. ‘A but not C’ keeps outcomes in A whose sum is more than 5.
2. Removing the 4 outcomes of A with sum ≤ 5 — (2,1), (2,2), (2,3), (4,1) — from A's 18 outcomes leaves 14.

AnswerA but not C — the 14 outcomes of A other than (2,1), (2,2), (2,3) and (4,1).

(vi) B or C

1. B or C is B ∪ C.
2. Take all 18 outcomes of B (first die odd), plus the outcomes of C not already in B: (2,1), (2,2), (2,3) and (4,1).

AnswerB or C — 22 outcomes: all of B plus those 4 extra outcomes from C.

(vii) B and C

1. B and C keeps outcomes with the first die odd AND sum ≤ 5.
2. Checking C's 10 outcomes, six of them have an odd first die.

AnswerB and C = {(1,1), (1,2), (1,3), (1,4), (3,1), (3,2)}.

(viii) A ∩ B′ ∩ C′

1. B′ is the same event as A (from part (ii)), so A ∩ B′ = A.
2. So A ∩ B′ ∩ C′ is just A ∩ C′, which is ‘A but not C’ from part (v).

AnswerA ∩ B′ ∩ C′ — the same 14 outcomes as part (v).

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(i) A and B are mutually exclusive

1. A is ‘first die even’ and B is ‘first die odd’.
2. The two can never happen on the same roll.

AnswerTrue — A and B are mutually exclusive.

(ii) mutually exclusive and exhaustive

1. Every first-die value is either even or odd, so A ∪ B is the whole sample space.
2. Together with A ∩ B = { }, this makes them exhaustive too.

AnswerTrue — A and B are mutually exclusive and exhaustive.

(iii) A = B′

1. B′ is ‘not odd on the first die’, which is exactly ‘even on the first die’.
2. That is the same event as A.

AnswerTrue — A and B′ are exactly the same 18 outcomes.

(iv) A and C are mutually exclusive

1. C contains outcomes such as (2,1), (2,2), (2,3) and (4,1).
2. All of these also have an even first die, so they lie in A too.

AnswerFalse — A and C share 4 outcomes, so they are not mutually exclusive.

(v) A and B′ are mutually exclusive

1. From part (iii), B′ is the same event as A.
2. So A and B′ share all 18 outcomes of A.

AnswerFalse — A and B′ are actually the same event, so they overlap completely.

(vi) A′, B′, C are mutually exclusive and exhaustive

1. A′ is the same as B, and B′ is the same as A.
2. A′ and C already overlap — for example at (1,1), which is odd on the first die and has sum ≤ 5.

AnswerFalse — A′ and C share outcomes such as (1,1), so the three are not mutually exclusive.

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