Suppose the probability that a softball player gets a hit in any single at-bat is 0.300. Assuming that her chance of getting a hit on a particular time at bat is independent of her other times at bat, what is the probability that she will not get a hit until her fourth time at bat in a game?

a.(43)(0.3)1(0.7)33051526=0.200=20.0%43(0.3)1(0.7)3

b.(43)(0.3)3(0.7)13051526=0.200=20.0%43(0.3)3(0.7)1

C.(41)(0.3)3(0.7)13051526=0.200=20.0%41(0.3)3(0.7)1

d.(0.3)3(0.7)13051526=0.200=20.0%(0.3)3(0.7)1

e.(0.3)1(0.7)33051526=0.200=20.0%(0.3)1(0.7)3

Short Answer

Expert verified

The answer is:

e.(0.3)1(0.7)3

Step by step solution

01

 Step 1: Given information 

We have to find the probability that she will not get a hit until her fourth time at bat in a game.

02

Explanation 

Geometrical probability:

P(X=k)=qk-1p=(1-p)k-1p

The probability is:

P(X=4)=(1-0.300)4-1(0.300)=(0.700)3(0.300)=(0.7)3(0.3)=(0.7)3(0.3)1=(0.3)1(0.7)3

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