Question

among 23 - to 28 - year - olds, 27% say they have driven a car while under the influence of alcohol. suppose five 23 - to 28 - year - olds are selected at random.
a. what is the probability that all five have driven a car while under the influence of alcohol?
b. what is the probability that at least one has not driven a car while under the influence of alcohol?

Explanation:

Step1: Calculate the probability for part a

The probability that a single 23 - to 28 - year - old has driven under the influence is \(p = 0.27\). Since the events are independent (selecting each person is an independent trial), the probability that all \(n=5\) have driven under the influence is given by the formula for the joint probability of independent events \(P(X = 5)=p\times p\times p\times p\times p\).

$$P(X = 5)=(0.27)^{5}$$

$$P(X = 5)=0.27\times0.27\times0.27\times0.27\times0.27$$

$$P(X = 5)=0.0014348907\approx0.0014$$

Step2: Calculate the probability for part b

First, find the probability that all 5 have driven under the influence. We know from part a that \(P(\text{all 5 have driven})=(0.27)^{5}\approx0.0014\).
The probability that at least one has not driven under the influence is the complement of the event that all 5 have driven under the influence.
Let \(A\) be the event that all 5 have driven under the influence. Then the probability that at least one has not driven under the influence, \(P(\text{at least one has not}) = 1 - P(A)\)

$$P(\text{at least one has not})=1-(0.27)^{5}$$

$$P(\text{at least one has not})=1 - 0.0014348907$$

$$P(\text{at least one has not})=0.9985651093\approx0.9986$$

Answer:

a. The probability that all five have driven a car while under the influence of alcohol is approximately \(0.0014\).
b. The probability that at least one has not driven a car while under the influence of alcohol is approximately \(0.9986\).