Example 29

EXAMPLE 29 Solving an “at least” problem

Using information in Example 28, find the probability that, in a random sample of 5 Americans ages 18–44, at least 1 of them smokes.

Solution

The phrase “at least” means that one or more of the five Americans smoke, so the probability we are looking for is:

$P (1 or 2 or 3 or 4 or 5 Americans smoke)$

Now, each of these events is mutually exclusive, meaning that our sample can’t yield exactly 1 American who smokes and exactly 2 Americans who smoke. Thus, by the Addition Rule for Mutually Exclusive Events, the above probability equals:

$P (1 American smokes) + P (2 Americans smoke) + \dots + P (5 Americans smoke)$

Calculating all these probabilities would take a while. So, we can use the probability of the complement we learned earlier to get us a shortcut. Note that “at least 1 American smokes” is the complement of “no Americans smoke.” Then, because the complement rule for probability is: $P (A^{C}) = 1 - P (A)$ , we get:

$\begin{matrix} P (At least 1 of the 5 Americans smokes) \\ = 1 - P (None of the 5 Americans smokes because the events are independent) \\ = 1 - P (1 st doesn' t smoke and 2 nd doesn' t smoke and \dots and 5 th doesn' t smoke) \\ = 1 - P (1 st doesn' t smoke) \cdot P (2 nd doesn' t smoke) \cdot \dots \cdot P (5 th doesn' t smoke) \\ = 1 - {(0.76)}^{5} because P (not smoking) = 0.76 \\ = 0.7464 \end{matrix}$

NOW YOU CAN DO

Exercises 101–104.