Question

In theHome Run Derby, each contestant is allowed to keep swinging the bat until they have made 10 “outs”. (An “out” is anything that is not a home run.) If a person has a 70% chance of hitting a home run on any given swing, what is the probability that he hits at least 10 home runs before his turn is up?

Answer 1

Let X be the number of trials for 10th out.

Then X is negative binomially distributed with p(success) = 0.3

P(X=x) = , x 10

We need to find P(X 20) because there will be at least 10 home runs in that case.

P(X 20) = 1 - P(X < 20)

where P(X < 20) = [ P(X=19) + P(X=18) +...............+P(X=10) ]

=> + +......................+