Porisma

Worked solution

Normalise the maximum score and count three-round orderings

An unbiased six-sided die is rolled twice. X is the higher value when they differ, or their common value when equal. Its distribution is P(X=r)=k(2r−1)P(X=r)=k(2r-1) for r=1,2,3,4,5,6.

(i) Find the exact value of k as a fraction in simplest form. marks

(ii) Find E(X)E(X). marks

One round consists of two rolls, with X as the score. A player plays three rounds.

(iii) Find the probability that the total score is 16. marks

[2017/TPJC/II/6]

CJC · Review problems · Question 2

Answer

k=1/36k=1/36; E(X)=161/36E(X)=161/36; total 16 probability 0.112 (3 s.f.).

(i) Sum every probability

∑r=16P(X=r)=1\sum_{r=1}^6P(X=r)=1, so k+3k+5k+7k+9k+11k=1k+3k+5k+7k+9k+11k=1. Therefore k=1/36k=1/36.

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(ii) Retain every weighted score

E(X)=1(k)+2(3k)+3(5k)+4(7k)+5(9k)+6(11k)=161k=161/36E(X)=1(k)+2(3k)+3(5k)+4(7k)+5(9k)+6(11k)=161k=161/36.

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(iii) Include permutations of both score patterns

The three-round total 16 requires scores 6,6,4 or 6,5,5 in any order. Each has 3!/2!3!/2! orders. Under the source independent-roll/round model, the probability is (11/36)2(7/36)(3!/2!)+(11/36)(9/36)2(3!/2!)=0.112(11/36)^2(7/36)(3!/2!)+(11/36)(9/36)^2(3!/2!)=0.112 to 3 significant figures. The source notation P(6,6,4)+P(6,5,5) groups all these orders.

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Common mistakes

For the three-round total, count all orders of each repeated-score pattern.

Solution notes

All three asks/printed marks retained; all normalisation and expectation terms and both three-round score patterns/permutation factors retained. Independent rolls and rounds are the source model.

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