Question

Suppose we roll a fair die four times. The probability that a $6$occurs on exactly one of the rolls is

(a) $4{\left(\frac{1}{6}\right)}^3{\left(\frac{5}{6}\right)}^1$

(b) $4{\left(\frac{1}{6}\right)}^1{\left(\frac{5}{6}\right)}^3$

(c) $6{\left(\frac{1}{6}\right)}^1{\left(\frac{5}{6}\right)}^3$

(d) ${\left(\frac{1}{6}\right)}^3{\left(\frac{5}{6}\right)}^1$

(e)${\left(\frac{1}{6}\right)}^1{\left(\frac{5}{6}\right)}^3$

Short answer

The probability that a $6$occurs on exactly one of the rolls is option (c) $4{\left(\frac{1}{6}\right)}^1{\left(\frac{5}{6}\right)}^3$.

Step-by-step solution

Given information

We roll a fair die four times.

Find the probability that a $6$occurs on exactly one of the rolls.

Explanation

The probability of $6$is $1/6$because all die rolls are independent

Then it follows the sampling distribution $(4,1/6)$

The probability formula is because of the binomial probability density.

$X~binomial(4,1/6)$

$P(X=1)=\left(\begin{array}{l}4\\ 1\end{array}\right){\left(\frac{1}{6}\right)}^1{\left(\frac{5}{6}\right)}^3=4{\left(\frac{1}{6}\right)}^1{\left(\frac{5}{6}\right)}^3$

The factorial component is

$\left(\begin{array}{l}4\\ 1\end{array}\right)=\frac{4!}{1!*(4-1)!}=4$

So the answer is (c).