Question

A basketball player has made 55% of his foul shots during the season. Assuming the shots are independent, what's the expected number of shots until he misses? The player is expected to take shots until the first miss. (Round to two decimal places as needed.)

Answer 1

Here are the answer to the question. Please let me know in case you've a question. .

1.

p = .55

Expected number of shorts until he misses is 1/p = 1/.55 = 1.82

Answer is 1.82

2. P(X=2) = ?

We hae been given parameters of binomial distribution. So, lets use them to solve the problem

P(X,n,p)
= P(2,15,.11)
= nCx*p^x * (1-p)^(n-x)
= 15C2 * .11^2 * .89^13
= 0.279