Question

Assume that when adults with smartphones are randomly​ selected, 53​% use them in meetings or classes. If 7 adult smartphone users are randomly​ selected, find the probability that exactly 5 of them use their smartphones in meetings or classes.

Answer 1

`P(X=5)=(7C5)(0.53)^5 (1-0.53)^(7-5)=0.194`

Then the probability that exactly 5 of them use their smartphones in meetings or classes is 0.194

Explanation:

Let X the random variable of interest "number of adults with smartphones", on this case we now that:

`X \sim Binom(n=7, p=0.53)`

The probability mass function for the Binomial distribution is given as:

`P(X)=(nCx)(p)^x (1-p)^(n-x)`

Where (nCx) means combinatory and it's given by this formula:

`nCx=(n!)/((n-x)! x!)`

And we want to find this probability:

`P(X=5)`

Using the probability mass function we got:

`P(X=5)=(7C5)(0.53)^5 (1-0.53)^(7-5)=0.194`

Then the probability that exactly 5 of them use their smartphones in meetings or classes is 0.194