Kason Wong

2022-12-27

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.

Toparellon12

Expert

Here there are 6 teams .
Out of this only 4 teams wins.
ie
total number of orders= 6P4=360
n(S)= 360
(1) There is only one way ABCD could be in sequence,
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
Let's create a sample set for this
There are six possible outcomes here with A as the first letter. 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

Do you have a similar question?