In Exercises 1–5, assume that 74% of randomly selected adults have a credit card (basedon results from an AARP Bulletin survey). Assume that a group of five adults is randomlyselected.

Find the probability that at least one of the five adults has a credit card. Does the result apply to five adult friends who are vacationing together? Why or why not?

The number of randomly selected adults are $n=5$.

The probability of randomly selected adults who have a credit card is $p=0.74$.

Let x represent the number of adults who have a credit card.

In the given scenario, the variable x will follow the binomial distribution.

The probability mass function of the binomial distribution is given as,

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

The probability that at least one of the five adults has a credit card is computed as,

$$\begin{gathered} P\left(x⩾1\right)=1-P\left(0\right) \\ =1-{}^5{C}_0{\left(0.74\right)}^0{\left(1-0.74\right)}^{5-0} \\ =1-\frac{5!}{0!\left(5-0\right)!}{\left(0.74\right)}^0{\left(0.26\right)}^5 \\ =0.9988 \\ =0.999 \end{gathered}$$

Thus, the probability that at least one of the five adults has a credit card is 0.999.

The probabilitythat at least one of the five adults has a credit card is 0.999 implies that the five friends are not randomly selected.

The financial situation of five friends is more likely to include credit cards as they are vacationing together.