Question

If Jeff gets 4 game pieces, what is the probability that he wins exactly 1 prize?

a. 0.25

b. 1.00

c. $\left(41\right)(0.25)3(0.75)3\left(\begin{array}{l}4\\ 1\end{array}\right)(0.25{)}^3(0.75{)}^3$

d. $\left(41\right)(0.25)1(0.75)3\left(\begin{array}{l}4\\ 1\end{array}\right)(0.25{)}^1(0.75{)}^3$

e.$(0.75{)}^3(0.75{)}^1$

Short answer

Option (d) is the correct answer

Step-by-step solution

Given Informaiton

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

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

Simplification

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

The probability that Jeff will win exactly 1 prize can be calculated as:

$$\begin{gathered} P(X=1)=\left(\begin{array}{l}4\\ 1\end{array}\right)(0.25{)}^1(1-0.25{)}^{4-1} \\ =\left(\begin{array}{l}4\\ 1\end{array}\right)(0.25{)}^1(1-0.25{)}^3 \end{gathered}$$

Thus, the required probability is $\left(\begin{array}{c}4\\ 1\end{array}\right)(0.25{)}^1(1-0.25{)}^3$.