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. 0.00430
b. 0.00467
c. 0.000358
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 $n$ 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, $n = 7$, $p=0.8$, and we want to find $P(X\lt3)=P(X = 0)+P(X = 1)+P(X = 2)$.

Step2: Calculate $P(X = 0)$

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

Step3: Calculate $P(X = 1)$

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

Step4: Calculate $P(X = 2)$

$C(7,2)=\frac{7!}{2!(7 - 2)!}=\frac{7\times6\times5!}{2\times1\times5!}=21$, $P(X = 2)=C(7,2)\times(0.8)^{2}\times(1 - 0.8)^{7 - 2}=21\times0.64\times(0.2)^{5}=21\times0.64\times0.00032 = 0.0043008$.

Step5: Calculate $P(X\lt3)$

$P(X\lt3)=P(X = 0)+P(X = 1)+P(X = 2)=0.0000128+0.0003584 + 0.0043008=0.004672$.

Answer:

B. 0.00467