find the probability of exactly three successes in six trials of a binomial experiment in which the…

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 successes, $p$ is the probability of success on a single trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$.

Step2: Identify the values of $n$, $k$, and $p$

Given that $n = 6$, $k = 3$, and $p=0.5$. First, calculate the combination $C(6,3)$. $C(6,3)=\frac{6!}{3!(6 - 3)!}=\frac{6!}{3!3!}=\frac{6\times5\times4\times3!}{3!\times3\times2\times1}=\frac{6\times5\times4}{3\times2\times1}=20$.

Step3: Calculate the probability

Substitute the values into the binomial - probability formula: $P(X = 3)=C(6,3)\times(0.5)^{3}\times(1 - 0.5)^{6 - 3}$. Since $1-0.5 = 0.5$, we have $P(X = 3)=20\times(0.5)^{3}\times(0.5)^{3}$. $P(X = 3)=20\times(0.5)^{6}$. $(0.5)^{6}=\frac{1}{64}$, and $20\times\frac{1}{64}=\frac{20}{64}=0.3125$. To convert to a percentage, multiply by 100: $0.3125\times100 = 31.25%$.

Answer:

$31.25%$