Question

In Exercises 21–24, assume that when adults with smartphones are randomly selected, 54% use them in meetings or classes (based on data from an LG Smartphone survey).

If 12 adult smartphone users are randomly selected, find the probability that fewer than 3 of them use their smartphones in meetings or classes.

Short answer

The probability of selecting less than three users who use their smartphones in meetings or classes is equal to 0.0095.

Step-by-step solution

Given information

A group of 12 adult smartphone users was selected. The probability of selecting a user who uses his/her smartphone in meetings or classes is equal to 0.54.

Required probability

Let X denote the users who use their smartphones in meetings or classes.

The number of trials (n) is given to be equal to 12.

The probability of success is calculated below:

$$\begin{gathered} p=54\% \\ =\frac{54}{100} \\ =0.54 \end{gathered}$$

The probability of failure is calculated below:

$$\begin{gathered} q=1-p \\ =1-0.54 \\ =0.46 \end{gathered}$$

The number of successes required in 12 trials should be less than three.

The binomial probability formula is as follows:

$P\left(X=x\right)={}^n{C}_x{\left(p\right)}^x{\left(q\right)}^{n-x}$

By using the binomial probability formula, the probability of getting less than three users who use their smartphones in meetings or classes is computed below:

$$\begin{gathered} P\left(X<3\right)=P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right) \\ ={}^{12}{C}_0{\left(0.54\right)}^0{\left(0.46\right)}^{12-0}+{}^{12}{C}_1{\left(0.54\right)}^1{\left(0.46\right)}^{12-1}+{}^{12}{C}_2{\left(0.54\right)}^2{\left(0.46\right)}^{12-2} \\ =0.0000897+0.0012644+0.0081641 \\ =0.0095 \end{gathered}$$

Therefore, the probability of getting fewer than three users who use their smartphones in meetings or classes is equal to 0.0095.