Question

assume that a procedure yields a binomial distribution with n = 5 trials and a probability of success of p = 0.05. use a binomial probability table to find the probability that the number of successes x is exactly 2. click on the icon to view the binomial probabilities table. p(2)= (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, $n = 5$, $k = 2$, and $p=0.05$, so $1-p = 0.95$.

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

$C(5,2)=\frac{5!}{2!(5 - 2)!}=\frac{5!}{2!3!}=\frac{5\times4\times3!}{2\times1\times3!}=10$.

Step3: Calculate the binomial probability

$P(X = 2)=C(5,2)\times(0.05)^{2}\times(0.95)^{3}=10\times0.0025\times0.857375 = 0.021434375$.

Step4: Round the result

Rounding $0.021434375$ to three decimal places, we get $0.021$.

Answer:

$0.021$