Chapter 10, Problem AP3.1CPT

To determine

The probability of the soccer player that she will not get a hit until her fourth time at bat in a game.

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

Answer to Problem AP3.1CPT

The correct option is ${\left(0.3\right)}^1{\left(0.7\right)}^3$ .

Explanation of Solution

Given:

Probability of the hit player: $0.300$

Concept Used:

Geometric Probability:

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

Calculation:

  $p:$ Probability of success: $0.300$

The first success among independent trials follows a geometric distribution.

So the interest in the probability that the first success occurs on the $4th$ bat, thus evaluate the definition of geometric probability at $k=4$

  $\begin{array}{l}P\left(X=4\right)={\left(1-0.300\right)}^{4-1}\left(0.300\right)\\ ={\left(0.700\right)}^3\left(0.300\right)\\ ={\left(0.7\right)}^3\left(0.3\right)\\ ={\left(0.7\right)}^3{\left(0.3\right)}^1\\ ={\left(0.3\right)}^1{\left(0.7\right)}^3\end{array}$

Conclusion:

Hence, the correct option is ${\left(0.3\right)}^1{\left(0.7\right)}^3$