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Assume that when adults with smartphones are randomly​ selected, 61% use them in meetings or classes. If 8 adult smartphone users are randomly​ selected, find the probability that exactly 3 of them use their smartphones in meetings or classes

 Feb 16, 2020
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Thanks for the A2A. I'm going to explain this very briefly. 

 

61% is 61/100 in fraction form. Now, we can see the probability of just 1 case. 

 

"find the probability that exactly 3 of them use their smartphones in meetings or classes". We can do \(\frac{61^3*39^5}{{100}^{8}}\) and get approximately .002. 

 

However, there are many cases so we will have to put in values for the combination formula.

 

After you substitute, you will get \(\frac{8!}{5!*3!}\). That is a total of 56 cases.

 

Multiply .00204 and 56. The answer in decimal form is  .11424.

 

In percent form, it is 11.424%. 

 

Cheers!

 Feb 16, 2020

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