Question

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?

$\text{a.}\left(43\right)(0.3)1(0.7)3\frac{305}{1526}=0.200=20.0\%\left(\begin{array}{l}4\\ 3\end{array}\right)(0.3{)}^1(0.7{)}^3$

$\text{b.}\left(43\right)(0.3)3(0.7)1\frac{305}{1526}=0.200=20.0\%\left(\begin{array}{l}4\\ 3\end{array}\right)(0.3{)}^3(0.7{)}^1$

$\text{C.}\left(41\right)(0.3)3(0.7)1\frac{305}{1526}=0.200=20.0\%\left(\begin{array}{c}4\\ 1\end{array}\right)(0.3{)}^3(0.7{)}^1$

$\text{d.}(0.3)3(0.7)1\frac{305}{1526}=0.200=20.0\%(0.3{)}^3(0.7{)}^1$

$\text{e.}(0.3)1(0.7)3^{\frac{305}{1526}}=0.200=20.0\%(0.3{)}^1(0.7{)}^3$

Short answer

The answer is:

e.$(0.3{)}^1(0.7{)}^3$

Step-by-step solution

 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.

Explanation 

Geometrical probability:

$P(X=k)=q^{k-1}p=(1-p{)}^{k-1}p$

The probability is:

$P(X=4)=(1-0.300{)}^{4-1}(0.300)=(0.700{)}^3(0.300)=(0.7{)}^3(0.3)=(0.7{)}^3(0.3{)}^1=(0.3{)}^1(0.7{)}^3$