Question

which expression represents the probability of rolling a 5 exactly three times in ten rolls of a number cube with six sides?
$10^{c}3\left(\frac{1}{6}\
ight)^{3}\left(\frac{1}{6}\
ight)^{7}$
$10^{c}3\left(\frac{1}{2}\
ight)^{3}\left(\frac{1}{2}\
ight)^{7}$
$10^{c}3\left(\frac{1}{6}\
ight)^{3}\left(\frac{5}{6}\
ight)^{7}$
$10^{c}3\left(\frac{1}{6}\
ight)^{7}\left(\frac{5}{6}\
ight)^{3}$

Explanation:

Step1: Recall binomial probability formula

The binomial - probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successful trials, $p$ is the probability of success in a single trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$.

Step2: Identify values for $n$, $k$, and $p$

In this case, the number of trials $n = 10$ (the number of rolls of the cube), the number of successful trials $k = 3$ (rolling a 5 exactly 3 times), and the probability of rolling a 5 in a single roll of a six - sided cube $p=\frac{1}{6}$. Then $1 - p = 1-\frac{1}{6}=\frac{5}{6}$.

Step3: Substitute values into the formula

Substituting $n = 10$, $k = 3$, and $p=\frac{1}{6}$ into the binomial - probability formula, we get $P(X = 3)=C(10,3)\times(\frac{1}{6})^{3}\times(\frac{5}{6})^{7}$, which can be written as $_{10}C_{3}(\frac{1}{6})^{3}(\frac{5}{6})^{7}$.

Answer:

$_{10}C_{3}(\frac{1}{6})^{3}(\frac{5}{6})^{7}$ (corresponding to the third option in the multiple - choice list)