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 :

Answer:

[tex]P(X=5)=(7C5)(0.53)^5 (1-0.53)^{7-5}=0.194[/tex]

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

Step-by-step explanation:

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

[tex]X \sim Binom(n=7, p=0.53)[/tex]

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

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]

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

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]

And we want to find this probability:

[tex] P(X=5)[/tex]

Using the probability mass function we got:

[tex]P(X=5)=(7C5)(0.53)^5 (1-0.53)^{7-5}=0.194[/tex]

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

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