Question

find the probability of exactly 5 successes in 6 trials of a binomial experiment in which the probability of success is 95%.
p = ?%
round to the nearest tenth of a percent

Explanation:

Step1: Recall 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 \(C(n,k)=\frac{n!}{k!(n - k)!}\)

Here, \(n = 6\), \(k = 5\), \(p=0.95\), and \(1 - p = 0.05\)

Step2: Calculate the combination \(C(6,5)\)

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Step3: Substitute values into the binomial formula

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Calculate \((0.95)^{5}=0.95\times0.95\times0.95\times0.95\times0.95 = 0.7737809375\)

Then \(P(X = 5)=6\times0.7737809375\times0.05\)

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Step4: Convert to percentage and round

Multiply by \(100\) to get percentage: \(P(X = 5)=0.23213428125\times100 = 23.213428125\%\)

Round to the nearest tenth of a percent: \(23.2\%\)

Answer:

\(23.2\%\)