Q. 5 - The Practice of Statistics for AP

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

Step-by-Step Solution

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 1: Given information

We roll a fair die four times. 

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

Step 2: 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).