Question

assume that a procedure yields a binomial distribution with n = 3 trials and a probability of success of p = 0.50. use a binomial probability table to find the probability that the number of successes x is exactly 1. click on the icon to view the binomial probabilities table. p(1)= (round to three decimal places as needed.)

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 $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $n = 3$, $k = 1$, and $p=0.5$.

Step2: Calculate the combination $C(n,k)$

$C(3,1)=\frac{3!}{1!(3 - 1)!}=\frac{3!}{1!2!}=\frac{3\times2!}{2!}=3$.

Step3: Calculate $(1 - p)^{n - k}$

Since $p = 0.5$, then $1-p=0.5$, and $n - k=3 - 1 = 2$. So, $(1 - p)^{n - k}=(0.5)^{2}=0.25$.

Step4: Calculate $p^{k}$

$p^{k}=(0.5)^{1}=0.5$.

Step5: Calculate the probability

$P(X = 1)=C(3,1)\times p^{1}\times(1 - p)^{3 - 1}=3\times0.5\times0.25 = 0.375$.

Answer:

$0.375$