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Worked solution

Compare a binomial majority with waiting for one further red pen

A manufacturer produces pens in two colors, red and blue. On average, 35% of the pens produced are red pens.

(i) Find the probability that in a sample of 10 pens, more than half of them are red pens. marks

(ii) The manufacturer offers a special price on red pens by selling them in packs of six. Pens are selected at random to fill each pack. Given that five red pens have been selected, find the least value of nn such that the probability that at most nn more pens need to be selected to complete a pack of 6 red pens is greater than 0.98. marks

[2007/AJC/II/8]

CJC · Review Problems · Question 6

Answer

(i) 0.09490.0949; (ii) least n=10n=10.

(i)

Let X∼B(10,0.35)X\sim B(10,0.35) count red pens. More than half means X≥6X\geq6. Then P(X>5)=1−P(X≤5)≈0.09493408017=0.0949P(X>5)=1-P(X\leq5)\approx0.09493408017=0.0949 to 3 significant figures.

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(ii)

Only one further red pen is needed. With independent selections of constant red probability 0.35, the probability that the first new red occurs by selection n is 0.35+0.35(0.65)+⋯+0.35(0.65)n−1=1−0.65n0.35+0.35(0.65)+\cdots+0.35(0.65)^{n-1}=1-0.65^n. Require 1−0.65n>0.981-0.65^n>0.98, so 0.65n<0.020.65^n<0.02 and n>ln⁡(0.02)/ln⁡(0.65)≈9.08n>\ln(0.02)/\ln(0.65)\approx9.08. Hence the least integer is n=10n=10. As an authored minimality check, n=9 gives about 0.979288 and n=10 about 0.986537.

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Common mistakes

n counts all additional pens drawn, including blue pens, until the final required red appears.

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