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Worked solution

A binomial model within a second binomial model

On average, 30% of the patients having a certain medical treatment experience severe headache. A random sample of 12 patients is given the treatment.

(i) Write down the distribution of the number of patients who experience severe headache, and state an assumption for the choice of the distribution.

(ii) Find the probability that fewer than 3 patients experience severe headache.

(iii) Five random samples of 12 patients are given the treatment. Find the probability that two of these five samples have fewer than 3 patients who experience severe headache.

[2007/ACJC/MYE/J2/3(Modified)]

CJC · Discussion · Question 4

Answer

(i) B(12,0.3)B(12,0.3) with the stated modelling assumption. (ii) 0.2530.253. (iii) 0.2670.267.

(i)

Let X be the number experiencing severe headache among 12. The model is X∼B(12,0.3)X\sim B(12,0.3). The source accepts either independence of patients’ headache outcomes or a constant headache probability throughout the sample as the stated assumption. The binomial calculation uses both independent outcomes and the common probability 0.3.

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(ii)

P(X<3)=P(X≤2)=0.252815347855…P(X<3)=P(X\leq2)=0.252815347855\ldots. The source gives 0.25282 (5 s.f.) and 0.253 (3 s.f.).

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(iii)

Let Y count samples, among five independent samples, having fewer than three affected patients. Then Y∼B(5,q)Y\sim B(5,q) with q=P(X≤2)q=P(X\leq2). The source uses q=0.25282. Keeping q unrounded, P(Y=2)=(52)q2(1−q)3=0.2666187564…P(Y=2)=\binom52q^2(1-q)^3=0.2666187564\ldots, so 0.267 (3 s.f.). The same final rounding results from the source’s five-significant-figure q.

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Common mistakes

The second model counts samples out of five, not patients out of sixty.

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