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.

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,

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

Hence, our required probability is 0.004672.

ummm seems that u didnt post the attachement !

answer: answer is b.

[m < b = 120º]

step-by-step explanation: