Question

a binomial probability experiment is conducted with the given parameters. compute the probability of x successes in the n independent trials of the experiment. n = 9, p = 0.2, x ≤ 3 the probability of x ≤ 3 successes is (round to four decimal places as needed.)

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

Step2: Calculate \(P(X = 0)\)

\(C(9,0)=\frac{9!}{0!(9-0)!}=1\)
\(P(X = 0)=1\times(0.2)^{0}\times(0.8)^{9}=0.1342\)

Step3: Calculate \(P(X = 1)\)

\(C(9,1)=\frac{9!}{1!(9 - 1)!}=9\)
\(P(X = 1)=9\times(0.2)^{1}\times(0.8)^{8}=0.3020\)

Step4: Calculate \(P(X = 2)\)

\(C(9,2)=\frac{9!}{2!(9-2)!}=36\)
\(P(X = 2)=36\times(0.2)^{2}\times(0.8)^{7}=0.3020\)

Step5: Calculate \(P(X = 3)\)

\(C(9,3)=\frac{9!}{3!(9 - 3)!}=84\)
\(P(X = 3)=84\times(0.2)^{3}\times(0.8)^{6}=0.1762\)

Step6: Sum up the probabilities

\(P(X\leq3)=P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)\)
\(P(X\leq3)=0.1342+0.3020 + 0.3020+0.1762=0.9144\)

Answer:

\(0.9144\)