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.

Answer 1

Answer: Our required probability is 0.004672.

Step-by-step explanation:

Since we have given that

Number of adults = 7

Probability of getting adult will watch prime time TV live = 0.8

We need to find the probability that fewer than 3 of the selected adults watch prime time live.

We will use "Binomial Distribution":

here, n = 7

p = 0.8

So, P(X<3)=P(X=0)+P(X=1)+P(X=2)

So, it becomes,

$P(X=0)=(1-0.8)^7=0.2^7=0.0000128$

$P(X=1)=^7C_1(0.8)(0.2)^6=0.0003584\\\\P(X=2)=^7C_2(0.8)^2(0.2)^5=0.0043$

So, probability that fewer than 3 of the selected adult watch prime time live is given by

$0.0000128+0.0003584+0.0043=0.004672$

Hence, our required probability is 0.004672.