If Jeff keeps playing until he wins a prize, what is the probability that he has to play the game exactly 5 times?

a. $(0.25{)}^5$

b. $(0.75{)}^4$

c. $(0.75{)}^5$

d. $(0.75{)}^4(0.25)$

e.$\left(51\right)(0.75)4(0.25)\left(\begin{array}{l}5\\ 1\end{array}\right)(0.75{)}^4(0.25)$

Number of trials $\left(n\right)=4$

Probability of success$\left(p\right)=\frac{1}{4}=0.25$

Consider, X is a random variable that depicts the number of games that are won follows the binomial distribution with n=4 and p=0.25.

The probability that Jeff will play exactly 5 times can be calculated as:

$\begin{array}{rcl}P(X& =& 5)=\left(\begin{array}{c}55\end{array}\right)(0.25{)}^5(1-0.25{)}^{5-5}\\ & =& 1\times (0.25{)}^5\times (1-0.25{)}^0\\ & =& 0.{25}^5\end{array}$

Thus, the required probability is $0.{25}^5$.

Hence, the correct option is (a).

(b) The probability that he has to play the game exactly 5 times will not be${\left(0.75\right)}^4$

(c) The probability that he has to play the game exactly 5 times will not be${\left(0.75\right)}^5$

(d) The probability that he has to play the game exactly 5 times will not be${\left(0.75\right)}^4\left(0.25\right)$

(e) The probability that he has to play the game exactly 5 times will not be $\left(\begin{array}{l}5\\ 1\end{array}\right)(0.75{)}^4(0.25)$