Question

question 4
the probability of a single customer purchasing an extended warranty is 0.35.
if there are 7 customers, what is the probability that 5 will purchase an extended warranty?
use binomial formula. show all work.
four decimal places
edit format table
12pt paragraph b i u a 2 t²

Explanation:

Step1: Identify the binomial formula parameters

The binomial formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(n = 7\) (number of customers), \(k = 5\) (number of customers purchasing warranty), \(p=0.35\) (probability of a single customer purchasing warranty), and \(1-p = 0.65\).
The combination formula \(C(n,k)=\frac{n!}{k!(n - k)!}\), so \(C(7,5)=\frac{7!}{5!(7 - 5)!}=\frac{7!}{5!2!}=\frac{7\times6}{2\times1}=21\).

Step2: Calculate the probability

Substitute the values into the binomial formula:
\(P(X = 5)=21\times(0.35)^{5}\times(0.65)^{2}\)
\((0.35)^{5}=0.35\times0.35\times0.35\times0.35\times0.35 = 0.0052521875\)
\((0.65)^{2}=0.4225\)
\(P(X = 5)=21\times0.0052521875\times0.4225\)
\(21\times0.0052521875 = 0.1102959375\)
\(0.1102959375\times0.4225=0.0465\) (rounded to four decimal places)

Answer:

\(0.0465\)