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Worked solution

Translate five burger events into probabilities

From past records, the purchases at a fast food outlet show that 63% of customers buy a burger. This figure is made up of 30% who buy a beef burger, 24% who buy a fish burger and 9% who buy a chicken burger. No one buys more than one burger. Find the probability that, in a random sample of 27 customers,

(i) at least 23 customers buy a burger,

(ii) less than 26 customers do not buy a fish burger,

(iii) more than 15 customers buy either a beef or a chicken burger,

(iv) no less than 8 customers do not buy a burger,

(v) exactly 4 customers buy a chicken burger given that fewer than 6 customers buy a chicken burger.

CJC · Discussion · Question 1

Answer

(i) 0.0108; (ii) 0.994; (iii) 0.0263; (iv) 0.839; (v) 0.136. All to 3 s.f.

(i)

Let X count burger buyers. X∼B(27,0.63)X\sim B(27,0.63). P(X≥23)=1−P(X≤22)=0.0107695942…=0.0108P(X\geq23)=1-P(X\leq22)=0.0107695942\ldots=\boxed{0.0108} (3 s.f.).

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(ii)

Let F count customers who do not buy a fish burger. F∼B(27,0.76)F\sim B(27,0.76). P(F<26)=P(F≤25)=0.994233770…=0.994P(F<26)=P(F\leq25)=0.994233770\ldots=\boxed{0.994} (3 s.f.).

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(iii)

Let Y count beef-or-chicken buyers. These categories are disjoint, so p=0.30+0.09=0.39. Y∼B(27,0.39)Y\sim B(27,0.39). P(Y>15)=1−P(Y≤15)=0.0263166800…=0.0263P(Y>15)=1-P(Y\leq15)=0.0263166800\ldots=\boxed{0.0263} (3 s.f.).

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(iv)

Let M count non-burger buyers. M∼B(27,0.37)M\sim B(27,0.37). P(M≥8)=1−P(M≤7)=0.839479284…=0.839P(M\geq8)=1-P(M\leq7)=0.839479284\ldots=\boxed{0.839} (3 s.f.).

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(v)

Let C count chicken-burger buyers. C∼B(27,0.09)C\sim B(27,0.09). Since the event C=4 is contained in C<6, P(C=4∣C<6)=P(C=4∩C<6)P(C<6)=P(C=4)P(C≤5)=0.135672557…=0.136P(C=4\mid C<6)=\dfrac{P(C=4\cap C<6)}{P(C<6)}=\dfrac{P(C=4)}{P(C\leq5)}=0.135672557\ldots=\boxed{0.136} (3 s.f.).

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Common mistakes

“Less than 26” means at most 25; “more than 15” starts at 16. For the conditional event, divide by the restricted sample probability.

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