Question

Assume that when adults with smartphones are randomly​ selected, 57​% use them in meetings or classes. If 9 adult smartphone users are randomly​ selected, find the probability that at least 3 of them use their smartphones in meetings or classes The probability is ?

Answer 1

Answer: 0.9617

Explanation:

Given : The proportion of adults use their phones in meetings or classes : p=0.57

Number of adults randomly selected : n= 9

Let x be the random variable that represents the number of adults use their phones in meetings or classes.

By using binomial probability formula, to find the probability of getting success in x trials.

`P(x)=^nC_xp^x(1-p)^(n-x)`

The probability that at least 3 of them use their smartphones in meetings or classes will be :-

`P(x\geq3)=1-P(x<3)\\\\=1-(P(0)+P(1)+P(2))\\\\=1-(^9C_0(0.57)^0(0.43)^9+^9C_1(0.57)^1(0.43)^8)\\\\=1-((0.43)^9+(9)(0.57)^1(0.43)^8+(36)(0.57)^2(0.43)^7)\\\\=0.961708368616\approx0.9617`

Hence, the required probability is 0.9617 .