Question

The starting line up for a basketball team is to consist of two forwards and three guards. Two brothers are on the team. Matthew is a forward and Tony a guard. There are four forwards and six guards from which to choose the line up. If the starting players are chosen at random, what is the probability that the two brothers will end up in the starting line up

Answer 1

0.25 = 25% probability that the two brothers will end up in the starting line up

Explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the players are chosen is not important, which means that the combinations formula is used to solve this question.

Combinations formula:

`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

Desired outcomes:

Matthew plus another forward from a set of 3.

Tony plus another two guards from a set of 5.

So

`D = C_(3,1)C_(5,2) = (3!)/(1!2!) * (5!)/(2!3!) = 3*10 = 30`

Total outcomes:

Two forwards from a set of 4.

Three guards from a set of 6.

So

`T = C_(4,2)C_(6,3) = (4!)/(2!2!) * (6!)/(3!3!) = 6*20 = 120`

What is the probability that the two brothers will end up in the starting line up?

`p = (D)/(T) = (30)/(120) = 0.25`

0.25 = 25% probability that the two brothers will end up in the starting line up