Question

find the probability of exactly one success in five trials of a binomial experiment in which the probability of success is 5%.
p = ?%
round to the nearest tenth of a percent

Explanation:

Step1: Identify the 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, and \(C(n,k)=\frac{n!}{k!(n - k)!}\)

Step2: Substitute the values into the formula

Here, \(n = 5\), \(k = 1\), \(p=0.05\), and \(1-p = 0.95\)

First, calculate \(C(5,1)=\frac{5!}{1!(5 - 1)!}=\frac{5!}{1!4!}=\frac{5\times4!}{4!}=5\)

Then, \(p^{k}=(0.05)^{1}=0.05\)

And \((1 - p)^{n - k}=(0.95)^{4}\)

\((0.95)^{4}=0.95\times0.95\times0.95\times0.95 = 0.81450625\)

Step3: Calculate the probability

\(P(X = 1)=C(5,1)\times p^{1}\times(1 - p)^{4}\)

\(P(X = 1)=5\times0.05\times0.81450625\)

\(P(X = 1)=5\times0.0407253125\)

\(P(X = 1)=0.2036265625\)

Step4: Convert to percentage and round

To convert to a percentage, multiply by \(100\): \(0.2036265625\times100 = 20.36265625\%\)

Rounding to the nearest tenth of a percent gives \(20.4\%\)

Answer:

\(20.4\%\)