Question

Assume that when adults with smartphones are randomly​ selected, ​44% use them in meetings or classes. If 8 adult smartphone users are randomly​ selected, find the probability that exactly 2 of them use their smartphones in meetings or classes.

Answer 1

Answer: The probability is 0.448

Step-by-step explanation:

First, 44% of adults use their smartphones in meetings or classes, then the probability that a random adult uses their smartphone on a meeting is:

p = 0.44

Then the probability that a random adult do not use it is:

q = 1 - 0.44 = 0.66

If 8 adult smartphone users are randomly​ selected, find the probability that exactly 2 of them use their smartphones in meetings or classes.

For a random (and fixed) case, the probability will be equal to the product of the individual probabilities for each person, this leads to:

P = (0.44)*(0.44)*(0.66)*(0.66)*(0.66)*(0.66)*(0.66)*(0.66)

P = (0.44)²*(0.66)⁶ = 0.016

And now we need to consider the different combinations. If we have C combinations, then the probability for the event will be C*P.

Remember that if we have N elements, the total number of different groups of K elements (from these N elements) is:

$C(N,K) = \frac{N!}{(N - K)!*K!}$

In this case, we have N = 8, and K = 2

Then we get:

$C(8, 2) = \frac{8!}{(8 - 2)!*2!} = \frac{8!}{6!*2!} = \frac{8*7}{2} = 4*7 = 28$

Then there are 28 different combinations.

This means that the probability is:

Probability = 28*P = 28*0.016 = 0.448

Probability = 0.448