Question

Please help with Part B

A basketball player gets 2 free-throw shots when she is fouled by a player on the opposing team. She misses the first shot 40% of the time. When she misses the first shot, she misses the second shot 5% of the time.

Part A: What is the probability of missing both free-throw shots?

Part B: What is the probability of making at least 1 free-throw shot?

Answer 1

The probability of missing both free-throw shots = 2%

The probability of making at least 1 free-throw shot = 98%

Explanation:

Given :

Probability that she misses the first shot = 40%

Probability she misses the second shot = 5%

Solution:

Part A: The probability of missing both free-throw shots:

The two shots are independent events so, we can use rule of independent events

`P(A \ \text{ and } B) = P(A) \cdot P(B)`

=>
`40\% * 5\%`

=>
`(40)/(100) * (5)/(100)`

=>
`0.4 * 0.05`

=>
`0.02`

=>
`2\%`

Part B: The probability of making at least 1 free-throw shot

Probability of making at least 1 free throw show shot = 1 - probability of miss both the shots

=>
`1 - 2\%`

=>
`1 - 0.02`

=>
`0.98`

=>
`98\%`