Question

39.8% of consumers believe that cash will be obsolete in the next 20 years. assume that 7 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. the probability is (round to three decimal places as needed )

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 on a single - trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $n = 7$, $p=0.398$, and $1 - p = 0.602$. We want to find $P(X\lt3)=P(X = 0)+P(X = 1)+P(X = 2)$.

Step2: Calculate $P(X = 0)$

$C(7,0)=\frac{7!}{0!(7 - 0)!}=1$. Then $P(X = 0)=C(7,0)\times(0.398)^{0}\times(0.602)^{7}=1\times1\times(0.602)^{7}\approx0.031$.

Step3: Calculate $P(X = 1)$

$C(7,1)=\frac{7!}{1!(7 - 1)!}=\frac{7!}{1!6!}=7$. Then $P(X = 1)=C(7,1)\times(0.398)^{1}\times(0.602)^{6}=7\times0.398\times(0.602)^{6}\approx0.131$.

Step4: Calculate $P(X = 2)$

$C(7,2)=\frac{7!}{2!(7 - 2)!}=\frac{7\times6}{2\times1}=21$. Then $P(X = 2)=C(7,2)\times(0.398)^{2}\times(0.602)^{5}=21\times(0.398)^{2}\times(0.602)^{5}\approx0.249$.

Step5: Calculate $P(X\lt3)$

$P(X\lt3)=P(X = 0)+P(X = 1)+P(X = 2)\approx0.031 + 0.131+0.249 = 0.411$.

Answer:

$0.411$