Fifteen people entered a cooking contest and the top 3 contestants win a cash prize. What is the percent chance any contestant wins a cash prize?

Answer 1

Consider that the probability of an event is given by:

(number of favourable cases)/(number of total cases)

The number of total cases, is the number of possible set of winners.

According to the text we can suppose that each of them has equal probability of get to the top 3.

Moreover, according to the text we do not consider the order of the top 3.

For this reason we are considering cominations of 3 element chosen among 15, that is

number of total cases = (15!) / [ (15-3)! 3! ] = 455

In order to count the number of favourable cases we can describe these cases as the set of winners when a particular winner is chosen, let's call it A.

If we call the other 2 winners B and C, then the set of the winners is

(A B C)

and to understand how many possible set we can get we have to understand in how many different ways B and C can be selected.

It is important to notice that B and C can be selected only among 14 possible winners becasue A cannot be chosen a second time.

For this reason, fixed A, the number of favourable cases is equivalent to choose 2 winners among 14 without considering the order, and so:

number of favourable cases = (14!) / [ (14-2)! 2! ] = 91

For this reason, the probability that we are looking for is 91/455 = 0.2

If you have missed any point of my explanation, just ask for more explanations :)