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.

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.

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.