Question

40.1% of consumers believe that cash will be obsolete in the next 20 years. assume that 8 consumers are randomly selected. find the probability that fewer than 3 of the selected consumers believe that cash will be obsolete in the next 20 years.

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 in a single trial, and \(C(n,k)=\frac{n!}{k!(n - k)!}\)

Here, \(n = 8\), \(p=0.401\), and we want to find \(P(X<3)=P(X = 0)+P(X = 1)+P(X = 2)\)

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

For \(k = 0\):
\(C(8,0)=\frac{8!}{0!(8 - 0)!}=1\)
\(P(X = 0)=C(8,0)\times(0.401)^{0}\times(1 - 0.401)^{8-0}=1\times1\times(0.599)^{8}\)
\(P(X = 0)\approx0.014\)

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

For \(k = 1\):
\(C(8,1)=\frac{8!}{1!(8 - 1)!}=8\)
\(P(X = 1)=C(8,1)\times(0.401)^{1}\times(0.599)^{7}\)
\(P(X = 1)=8\times0.401\times(0.599)^{7}\approx8\times0.401\times0.023\approx0.074\)

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

For \(k = 2\):
\(C(8,2)=\frac{8!}{2!(8 - 2)!}=\frac{8\times7}{2\times1}=28\)
\(P(X = 2)=C(8,2)\times(0.401)^{2}\times(0.599)^{6}\)
\(P(X = 2)=28\times0.160801\times(0.599)^{6}\approx28\times0.160801\times0.038\approx0.171\)

Step5: Sum the probabilities

\(P(X<3)=P(X = 0)+P(X = 1)+P(X = 2)\)
\(P(X<3)\approx0.014 + 0.074+0.171=0.259\)

Answer:

\(0.259\)