Assume that when adults with smartphones are randomly selected, 49% 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.

Using the binomial distribution, it is found that there is a 0.8999 = 89.99% probability that at least 3 of them use their smartphones in meetings or classes.

For each adult, there are only two possible outcomes. Either they use their smartphones in meetings or classes, or they do not. The probability of an adult using their smartphones in meetings or classes is independent of any other adult, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

In this problem:

The desired probability is:

In which:

Then

0.8999 = 89.99% probability that at least 3 of them use their smartphones in meetings or classes.

A similar problem is given at link

0.18173219

The probability is that at least 6 of 9 adults use their smartphones in meetings or classes is 0.18173219

Hope this helps ;)

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x