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find the probability of exactly 6 successes in 7 trials of a binomial e…

Question

find the probability of exactly 6 successes in 7 trials of a binomial experiment in which the probability of success is 80%. 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, \(n = 7\), \(k=6\), \(p = 0.8\), and \(1-p=0.2\)

First, calculate \(C(7,6)=\frac{7!}{6!(7 - 6)!}=\frac{7!}{6!1!}=\frac{7\times6!}{6!×1}=7\)

Then, \(p^{k}=(0.8)^{6}\) and \((1 - p)^{n - k}=(0.2)^{1}\)

\(P(X = 6)=C(7,6)\times(0.8)^{6}\times(0.2)^{1}\)

\(P(X = 6)=7\times0.262144\times0.2\)

Step3: Calculate the result

\(7\times0.262144\times0.2 = 7\times0.0524288=0.3670016\)

To convert to a percentage, multiply by \(100\): \(0.3670016\times100 = 36.70016\%\)

Rounding to the nearest tenth of a percent gives \(36.7\%\)

Answer:

\(36.7\%\)