Answer to Question #159239 in Statistics and Probability for Kendra

Question

There are 12 people waiting for extra tickets to a comedy show. Three tickets become available and are randomly assigned. How many different groupings of people can be given the tickets? What is the probability you and your two friends are the 3 people that receive the tickets?

Answer 1

The first person (from 12) to get a ticket can be selected in 12 different ways, then the second person (from the remaining 11) to get a ticket can be selected in 11 different ways and the third one can be selected in 10 different ways.

Therefore, the total number of ways to form an ordered 3-sequence of persons from the group of 12 people is equal to the product 12*11*10 = 1320.

But any permutation of the order in which the selection is made gives the same result. The total number of permutations on the set of 3 elements is equal to 3! = 3*2*1 = 6. Hence, any group of 3 persons gives us 6 different ordered 3-sequences of persons.

The total number of different groups of 3 persons selected from 12 people is equal to 1320/6 = 220.

If one suppose that all 3-groups can be selected with equal probability, then the probability that me and my two friends are the 3 people that receive the tickets is equal to $1/220\approx0.45\%$.