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Miscellaneous Examples · Example 12

Q.In a relay race there are five teams A, B, C, D and E.

(a) What is the probability that A, B and C finish first, second and third, respectively?
(b) What is the probability that A, B and C are first three to finish (in any order)? (Assume that all finishing orders are equally likely.)
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All 5!=1205!=120 finishing orders of the five teams are equally likely. (a) Exactly one order has A, B, C first, second, third with D, E free in 2!=22!=2 ways, giving 2120=160\dfrac{2}{120}=\dfrac{1}{60}. (b) A, B, C fill the top three in any order in 3!×2!=123!\times2!=12 ways, giving 12120=110\dfrac{12}{120}=\dfrac{1}{10}.

Since all finishing orders are equally likely, we use classical probability:

P=favourable orderstotal orders.P=\dfrac{\text{favourable orders}}{\text{total orders}}.

Five teams can finish in 5!=1205!=120 different orders, so the denominator is 120120.

Part (a): A first, B second, C third

Fix A in 1st, B in 2nd and C in 3rd. The remaining teams D and E fill the last two places in 2!=22!=2 ways, so there are 22 favourable orders:

P(a)=2120=160.P(a)=\dfrac{2}{120}=\dfrac{1}{60}.

Part (b): A, B, C are the first three (any order) …

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