Question

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

$$\begin{gathered} a.4\left(16\right)3\left(56\right)14{\left(\frac{1}{6}\right)}^3{\left(\frac{5}{6}\right)}^1 \\ b.\left(16\right)3\left(56\right)1{\left(\frac{1}{6}\right)}^3{\left(\frac{5}{6}\right)}^1 \\ c.4\left(16\right)1\left(56\right)3{\left(\frac{1}{6}\right)}^1{\left(\frac{5}{6}\right)}^3 \\ d.\left(16\right)1\left(56\right)3{\left(\frac{1}{6}\right)}^1{\left(\frac{5}{6}\right)}^3 \\ e.6\left(16\right)1\left(56\right)3^6{\left(\frac{1}{6}\right)}^1{\left(\frac{5}{6}\right)}^3 \end{gathered}$$

Short answer

The probability that a 6 occurs on exactly one of the rolls is$4\left(16\right)1\left(56\right)3{\left(\frac{1}{6}\right)}^1{\left(\frac{5}{6}\right)}^3$.

Step-by-step solution

Given Information

Let roll four times with a fair die

Formula used:

$P(X=r)={}^{n}C_1\left(p{)}^r\right(q{)}^{n-r}$

Explanation for correct option

Because each die roll is unique,

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

$P(X=1)={}^{4}C_1{\left(\frac{1}{6}\right)}^1{\left(\frac{5}{6}\right)}^3$

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

Therefore, the correct option is (c)

Explanation for incorrect option

(a) The probability that a 6 occurs on exactly one of the rolls is not $4\left(16\right)3\left(56\right)14{\left(\frac{1}{6}\right)}^3{\left(\frac{5}{6}\right)}^1$

(b) The probability that a 6 occurs on exactly one of the rolls is not $\left(16\right)3\left(56\right)1{\left(\frac{1}{6}\right)}^3{\left(\frac{5}{6}\right)}^1$

(d) The probability that a 6 occurs on exactly one of the rolls is not $\left(16\right)1\left(56\right)3{\left(\frac{1}{6}\right)}^1{\left(\frac{5}{6}\right)}^3$

(e) The probability that a 6 occurs on exactly one of the rolls is not $6\left(16\right)1\left(56\right)3^6{\left(\frac{1}{6}\right)}^1{\left(\frac{5}{6}\right)}^3$