Question

a fair die is rolled four times. a 4 is considered \success,\ while all other outcomes are \failures.\ find the probability of 3 successes.
the probability of 3 successes is .
(type an integer or a simplified fraction.)

Explanation:

Step1: Identify probability of single - success and failure

The probability of getting a 4 (success) in a single roll of a fair die is $p=\frac{1}{6}$, and the probability of not getting a 4 (failure) is $q = 1 - p=1-\frac{1}{6}=\frac{5}{6}$.

Step2: Use binomial probability formula

The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times q^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $n = 4$, $k = 3$. First, calculate the combination $C(4,3)=\frac{4!}{3!(4 - 3)!}=\frac{4!}{3!1!}=\frac{4\times3!}{3!×1}=4$.

Step3: Calculate the probability

Substitute $n = 4$, $k = 3$, $p=\frac{1}{6}$, and $q=\frac{5}{6}$ into the binomial - probability formula: $P(X = 3)=C(4,3)\times(\frac{1}{6})^{3}\times(\frac{5}{6})^{4 - 3}=4\times\frac{1}{216}\times\frac{5}{6}=\frac{20}{1296}=\frac{5}{324}$.

Answer:

$\frac{5}{324}$