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Worked solution

Distinguish not-all-together from pairwise separation

A group of fifteen people consists of one pair of sisters, one set of three brothers and ten other people. They are arranged randomly in a line.

(i) Find the probability that the sisters are next to each other.

(ii) Find the probability that the brothers are not all next to one another.

(iii) Find the probability that the sisters are next to each other and the brothers are all next to one another.

(iv) Find the probability that either the sisters are next to each other or the brothers are all next to one another or both.

Instead the fifteen people are arranged in a circle.

(v) Find the probability that the sisters are next to each other.

[2012/II/7]

CJC · Discussion · Question 4

Answer

(i) 2/152/15; (ii) 34/3534/35; (iii) 2/4552/455; (iv) 43/27343/273; (v) 1/71/7.

Use 15! equally likely linear arrangements

The fifteen people are distinct, so the line has 15!15! possible arrangements. Internal orders of family blocks must still be counted.

How did this part go?

(i) Group the sisters

The sisters form one block beside the ten others and three individual brothers:14 objects. There are 14! 2!14!\,2! favourable arrangements, including the sisters’ two orders. Thus P(S)=14! 2!/15!=2/15P(S)=14!\,2!/15!=2/15.

How did this part go?

(ii) Exclude a single three-brother block

If all brothers are together, their block, two sisters and ten others make 13 objects, with 13! 3!13!\,3! arrangements. Therefore P(brothers not all together)=1−13! 3!/15!=34/35P(\text{brothers not all together})=1-13!\,3!/15!=34/35. This allows two brothers to be adjacent; the ask is not pairwise separation.

How did this part go?

(iii) Group each family separately

The sisters’ block, brothers’ block and ten others are 12 objects. Their internal orders contribute 2!2! and 3!3!. Thus P(S∩B)=12! 3! 2!/15!=2/455P(S\cap B)=12!\,3!\,2!/15!=2/455, where B denotes all brothers together.

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(iv) Include either family block

With S as above and B all brothers together, P(S∪B)=P(S)+P(B)−P(S∩B)=2/15+(1−34/35)−2/455=43/273P(S\cup B)=P(S)+P(B)-P(S\cap B)=2/15+(1-34/35)-2/455=43/273. The overlap must be removed once.

How did this part go?

(v) Change both counts for the circle

The total number is (15−1)!=14!(15-1)!=14!. With sisters together there are 14 circular objects, hence (14−1)! 2!=13! 2(14-1)!\,2!=13!\,2 favourable arrangements. The probability is 13! 2/14!=1/713!\,2/14!=1/7.

How did this part go?

Common mistakes

Not all three brothers together does not mean no two brothers are adjacent. A circle also needs a different denominator from a line.

Solution notes

All printed asks, working, probability calculations and source annotations visually compared and retained. No marks are printed for this question.

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