Question

. A player in Boise State University basketball team makes 80% of his free throws. At the end of a game with the basketball team of Portland State University, his team is losing by one point. He is fouled attempting a three-point shot and is awarded two free throws because of the aggressive defense violation from a PSU player. Suppose that each free throw is independent, and the game is over right after the two free throws. What is the probability that BSU win this game

Answer 1

64% probability that BSU win this game

Explanation:

Answer 2

64% probability that BSU win this game

Explanation:

For each free throw, there are only two possible outcomes. Either he makes it, or he does not. The probability of making a free throw is independent of other free throws. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`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)!)`

And p is the probability of X happening.

Two free throws.

This means that
`n = 2`

A player in Boise State University basketball team makes 80% of his free throws.

This means that
`p = 0.8`

What is the probability that BSU win this game

The team is trailing by 1 and he will attemp 2 free throws. If he makes both, BSU wins the game.

So we have to find P(X = 2).

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 2) = C_(2,2).(0.8)^(2).(0.2)^(0) = 0.64`

64% probability that BSU win this game