Question

A baseball team consists of 14 American-born players and 11 foreign-born players. Nine players must be selected to start the game. Find the probability (to 4 decimal places) of selecting 5 American-born players and 4 foreign-born players

Answer 1

32.33%

Explanation:

There is no marking that can differentiate the position or the player, so both of their order is not important. Since the order not important, we use combination instead of permutation.

There are 25 players and you can choose 9 of them. The number of ways to do this will be: 9C25= 25! / (9! * (25-9)! ) = 2042975 ways

You want to choose 5 American-born out of 14 American-born and 4 foreign-born out of 11 foreign-born. The ways to do that will be: 14C5 + 4C11=

14! / (5! * (14-5)! ) * 11! / (4! * (11-4)! ) =2002 * 330 = 660660 ways

The probability will be: 660660/2042975 = 0.3233= 32.33%