which expression represents the probability of rolling a 5 exactly three times in ten rolls of a number cube…

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

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

Answer

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 $p$

Here, $n = 10$ (the number of rolls of the number - cube), $k = 3$ (the number of times we want to roll a 5), and $p=\frac{1}{6}$ (the probability of rolling a 5 on a single roll of a six - sided number cube). Then $1-p = 1-\frac{1}{6}=\frac{5}{6}$.

Step3: Substitute values into the formula

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

Answer:

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