In a relay race,there are six teams A,B,C,D,E,F.(i)What is the probability that A,B,C,D finish first, second, third and fourth respectively.(ii)What is the probability that A,B,C and D are first four to finish(in any order).Assume that all finishing orders are equally likely.

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Solution

Here there are 6 teams .
Out of this only 4 teams wins.
ie
total number of orders= 6P4=360
n(S)= 360

(1) For ABCD to come in order, there is only one possibility ,
ie n(L) = 1

there for probability of A,B,C,D finish first, second, third and fourth respectively= n(S)/n(L)
= 1/360

(2) probability that A,B,C and D are first four to finish
lets write the sample set for this
M= ABCD, ABDC,ACBD,ACDB,ADBC,ADCB,
Here with A as first letter we could write 6 outcomes. Simillarly with BCD there are 6 outcomes each,
total number of such outcomes= 6*4=24

therefor probability that A,B,C and D are first four to finish = 24/360
=1/15

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