Question

Assume that when adults with smartphones are randomly​ selected, 47​% use them in meetings or classes. If 10 adult smartphone users are randomly​ selected, find the probability that fewer than 4 of them use their smartphones in meetings or classes.

Answer 1

Answer:

22.51% probability that fewer than 4 of them use their smartphones in meetings or classes.

Step-by-step explanation:

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 phone is independent of other adults. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening

47​% use them in meetings or classes.

This means that $p = 0.47$

10 adult smartphone users are randomly​ selected

This means that $n = 10$

Find the probability that fewer than 4 of them use their smartphones in meetings or classes.

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.47)^{0}.(0.53)^{10} = 0.0017$

$P(X = 1) = C_{10,1}.(0.47)^{1}.(0.53)^{9} = 0.0151$

$P(X = 2) = C_{10,2}.(0.47)^{2}.(0.53)^{8} = 0.0619$

$P(X = 3) = C_{10,3}.(0.47)^{3}.(0.53)^{7} = 0.1464$

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0017 + 0.0151 + 0.0619 + 0.1464 = 0.2251$

22.51% probability that fewer than 4 of them use their smartphones in meetings or classes.