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Worked solution

Choose hypergeometric or binomial sampling from replacement rules

A jar contains 10 blue and 8 red marbles. Five marbles are randomly drawn from the box, one by one and without replacement.

(i) Explain why it is inappropriate to model the number of blue marbles by a binomial distribution. mark

(ii) Find the probability that exactly three marbles are blue. marks

Another jar contains 20 blue and 12 red marbles. nn marbles are randomly drawn from the box, one by one and with replacement. The number of red marbles drawn is denoted by RR.

(iii) Given that the mean of RR is 4.5, find nn and P(R>4)P(R>4). marks

(iv) Given instead that P(R=0 or 1)<0.01P(R=0\text{ or }1)<0.01, write down an inequality for nn and find the least value of nn. marks

[2018/JJC/II/8]

CJC · Review Problems · Question 15

Answer

(ii) 20/51≈0.39220/51\approx0.392; (iii) n=12n=12, P(R>4)=0.490P(R>4)=0.490; (iv) (5/8)n+n(3/8)(5/8)n−1<0.01(5/8)^n+n(3/8)(5/8)^{n-1}<0.01, least n=15n=15.

(i)

Without replacement from this small fixed jar, the colour probabilities conditional on preceding draws change and the draws are dependent. A binomial model is therefore inappropriate.

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(ii)

Both printed routes are retained. Counting selections gives (103)(82)(185)=2051≈0.392\dfrac{\binom{10}3\binom82}{\binom{18}5}=\dfrac{20}{51}\approx0.392 (3 s.f.). Alternatively, choose one order with three blues followed by two reds and multiply by the number of colour orders: 10189178168157145!3!2!=2051\dfrac{10}{18}\dfrac9{17}\dfrac8{16}\dfrac8{15}\dfrac7{14}\dfrac{5!}{3!2!}=\dfrac{20}{51}. This is an exact without-replacement calculation.

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(iii)

With replacement, R∼B(n,12/32)=B(n,3/8)R\sim B(n,12/32)=B(n,3/8). The mean gives (3/8)n=4.5(3/8)n=4.5, so n=12n=12. Then P(R>4)=1−P(R≤4)≈0.4897166289=0.490P(R>4)=1-P(R\leq4)\approx0.4897166289=0.490 (3 s.f.).

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(iv)

This is a new condition, not n=12 from (iii). Require (5/8)n+n(3/8)(5/8)n−1<0.01(5/8)^n+n(3/8)(5/8)^{n-1}<0.01. The least integer is n=15. An authored adjacent-value check gives .01304512054 at n=14 and .00867361738 at n=15. The event of at most one red is decreasing with n, so this establishes minimality.

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Common mistakes

The word “instead” in (iv) replaces the mean condition from (iii).

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