Question

find the probability of exactly 3 successes in 6 trials of a binomial experiment in which the probability of success is 75%.
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, \(n = 6\), \(k = 3\), \(p=0.75\), and \(1-p = 0.25\)

First, calculate \(C(6,3)=\frac{6!}{3!(6 - 3)!}=\frac{6!}{3!3!}=\frac{6\times5\times4}{3\times2\times1}=20\)

Then, \(p^{k}=(0.75)^{3}=0.421875\)

And \((1 - p)^{n - k}=(0.25)^{3}=0.015625\)

Step3: Calculate the probability

\(P(X = 3)=C(6,3)\times(0.75)^{3}\times(0.25)^{3}\)

\(P(X = 3)=20\times0.421875\times0.015625\)

\(P(X = 3)=20\times0.006591796875\)

\(P(X = 3)=0.1318359375\)

Step4: Convert to percentage and round

To convert to a percentage, multiply by \(100\): \(P(X = 3)=0.1318359375\times100 = 13.18359375\%\)

Rounding to the nearest tenth of a percent gives \(13.2\%\)

Answer:

\(13.2\%\)