Please show work thank you A college basketball player has p

Please show work, thank you.

A college basketball player has probability 5/6 of making his free-throw shots, and free-throw shots are independent. What is the probability that the basketball player will make at least 5 of his next 7 free-throw shots? (Use the Binomial Distribution - Show your work.)

Solution

using binomial distribution

P( X=x) = nCx * p^x * q^(n-x)

n = 7

p = 5/6

probability that atleast make 5 shots

= P(X >=5)

= P(X=5) + P(X=6) + P(X=7)

= 7C5 * (5/6)^5 * (1/6)^2 + 7C6 (5/6)^6 * (1/6) + 7C7 * (5/6)^7

= 0.9042