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assume that a procedure yields a binomial distribution with a trial rep…

Question

assume that a procedure yields a binomial distribution with a trial repeated n = 30 times. use the binomial probability formula to find the probability of x = 5 successes given the probability p = 1/5 of success on a single trial. round to three decimal places.
a. 0.067
b. 0.172
c. 0.421
d. 0.198

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 \(C(n,k)=\frac{n!}{k!(n - k)!}\), \(n = 30\), \(k = 5\), \(p=\frac{1}{5}=0.2\), and \(1-p = 0.8\).

Step2: Calculate combination \(C(n,k)\)

\(C(30,5)=\frac{30!}{5!(30 - 5)!}=\frac{30!}{5!×25!}=\frac{30\times29\times28\times27\times26}{5\times4\times3\times2\times1}=142506\)

Step3: Calculate \(p^{k}\) and \((1 - p)^{n - k}\)

\(p^{k}=(0.2)^{5}=0.00032\), \((1 - p)^{n - k}=(0.8)^{25}\approx0.00377789\)

Step4: Calculate \(P(X = 5)\)

\(P(X = 5)=C(30,5)\times p^{5}\times(0.8)^{25}=142506\times0.00032\times0.00377789\approx0.172\)

Answer:

B. \(0.172\)