Question

based on a comcast survey, there is a 0.8 probability that a randomly selected adult will watch prime - time tv live, instead of online, on dvr, etc. assume that seven adults are randomly selected. find the probability that fewer than three of the selected adults watch prime - time live.
a. 0.00055
b. 0.00467
c. 0.00430
d. 0.0512

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 \(n = 7\) (number of trials, i.e., number of adults), \(p=0.8\) (probability of an adult watching prime - time live), and we want to find \(P(X<3)=P(X = 0)+P(X = 1)+P(X = 2)\)

Step2: Calculate \(C(n,k)=\frac{n!}{k!(n - k)!}\)

• For \(k = 0\):

\(C(7,0)=\frac{7!}{0!(7-0)!}=1\)
\(P(X = 0)=C(7,0)\times(0.8)^{0}\times(1 - 0.8)^{7-0}=1\times1\times(0.2)^{7}=0.0000128\)

• For \(k = 1\):

\(C(7,1)=\frac{7!}{1!(7 - 1)!}=\frac{7!}{1!6!}=7\)
\(P(X = 1)=C(7,1)\times(0.8)^{1}\times(0.2)^{6}=7\times0.8\times0.000064 = 0.0003584\)

• For \(k = 2\):

\(C(7,2)=\frac{7!}{2!(7-2)!}=\frac{7\times6\times5!}{2\times1\times5!}=21\)
\(P(X = 2)=C(7,2)\times(0.8)^{2}\times(0.2)^{5}=21\times0.64\times0.00032=0.0043008\)

Step3: Sum up the probabilities

\(P(X<3)=P(X = 0)+P(X = 1)+P(X = 2)=0.0000128+0.0003584 + 0.0043008=0.004672\approx0.00467\)

Answer:

B. \(0.00467\)