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Worked solution

Find a repeat-draw distribution and approximate its sample mean

A bag has three identically shaped balls numbered 1,2,3. Draw one at a time at random with replacement. Let XX be the number of draws until any ball is drawn a second time; matching draws need not be consecutive.

(i) Show P(X=4)=2/9P(X=4)=2/9 and find the distribution of XX. marks

(ii) Show E(X)=26/9E(X)=26/9 and find exact Var⁡(X)\operatorname{Var}(X). marks

(iii) Using a suitable approximation, find the probability that the mean of 44 independent observations exceeds 3. marks

[2017/NJC/II/7]

CJC · Review problems · Question 9

Answer

(i) Probabilities 1/3,4/9,2/91/3,4/9,2/9 at 2,3,4. (ii) E(X)=26/9E(X)=26/9, variance 44/8144/81. (iii) CLT approximation 0.1590.159.

(i) Enumerate the first repeated draw

The possible values are 2,3,4. We have P(X=2)=1/3P(X=2)=1/3, P(X=3)=(2/3)(2/3)=4/9P(X=3)=(2/3)(2/3)=4/9, and P(X=4)=(3/3)(2/3)(1/3)(3/3)=2/9P(X=4)=(3/3)(2/3)(1/3)(3/3)=2/9: the first three draws must all differ, after which a repeat is certain.

xx234
P(X=x)P(X=x)1/31/34/94/92/92/9
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(ii) Exact first and second moments

E(X)=2/3+3(4/9)+4(2/9)=26/9E(X)=2/3+3(4/9)+4(2/9)=26/9. Also E(X2)=4/3+9(4/9)+16(2/9)=80/9E(X^2)=4/3+9(4/9)+16(2/9)=80/9. Hence Var⁡(X)=80/9−(26/9)2=44/81\operatorname{Var}(X)=80/9-(26/9)^2=44/81.

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(iii) Printed CLT approximation

For 44 independent observations, Xˉ∼˙N(26/9,(44/81)/44)=N(26/9,1/81)\bar X\mathrel{\dot\sim}N(26/9,(44/81)/44)=N(26/9,1/81). Standardising the threshold 3 gives Z>1Z>1, so P(Xˉ>3)≈1−Φ(1)=0.159P(\bar X>3)\approx1-\Phi(1)=0.159, to 3 s.f.

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Common mistakes

The question requests an approximation for a discrete mean. The strict event is not an exact continuous-normal tail.

Solution notes

Independent authored check: coefficients of (3+4t+2t²)^44 give the exact strict total event sumX≥133, probability 0.1354420153. The printed answer 0.159 is retained as the requested unadjusted CLT approximation.

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