Question

At Least One. In Exercises 5–12, find the probability.

Wi-Fi Based on a poll conducted through the e-edition of USA Today, 67% of Internet users are more careful about personal information when using a public Wi-Fi hotspot. What is the probability that among four randomly selected Internet users, at least one is more careful about personal information when using a public Wi-Fi hotspot? How is the result affected by the additional information that the survey subjects volunteered to respond?

Short answer

The probability of getting at least one user who is more careful about personal information when using public Wi-Fi hotspot is equal to 0.988.

As the users were not selected at random and responded voluntarily, the results can be greatly affected and might be incorrect.

Step-by-step solution

Given information

The probability of selecting a user who is more careful about personal information when using public Wi-Fi hotspot is 0.67.

Four users are selected at random.

Define the event for “at least one”

The probability of an event occurring "at least once" is the likelihood of a situation occurring once or more than once.

It is also equal to one minus the chance that the event will not occur at all.

$P\left(\text{A}\text{happening}\text{at}\text{least}\text{once}\right)=1-P\left(\text{not}\text{A}\right)$

Compute the probability of the event

Let A be the event of selecting a user who is careful about personal information when using public Wi-Fi hotspot.

Then, the complementary event would be selecting a user who is not very careful about personal information when using public Wi-Fi hotspot.

The probability of selecting such a user is given by:

$$\begin{gathered} P\left(A\right)=\frac{67}{100} \\ =0.67 \end{gathered}$$

Thus, P(A) is 0.67.

The probability of selecting a user who is not very careful about personal information when using public Wi-Fi hotspot is equal to:

$$\begin{gathered} P\left(\overline{A}\right)=1-P\left(A\right) \\ =1-0.67 \\ =0.33 \end{gathered}$$

Thus,$P\left(\overline{A}\right)$ is equal to 0.33.

Four users are selected at random.

The probability of getting all four users who are not careful about personal information when using public Wi-Fi hotspot is given by:

$$\begin{gathered} P\left(\text{all}\text{good}\right)=\left(0.33\right)\times \left(0.33\right)\times \left(0.33\right)\times \left(0.33\right) \\ =0.0119 \end{gathered}$$

The probability of getting at least one careful user is one minus the probability of getting no careful user:

$$\begin{gathered} P\left(\text{at}\text{least}\text{1}\text{careful}\text{user}\right)=1-P\left(\text{no}\text{careful}\text{user}\right) \\ =1-0.0119 \\ =0.988 \end{gathered}$$

Therefore, the probability of getting at least one who is not careful about personal information when using public Wi-Fi hotspot is equal to 0.988.

Discuss the validity of the result

It is known that the data is based on the information collected through the e-edition of USA Today. The users voluntarily responded.

As the users responded voluntarily and were not chosen randomly, the results will be considerably influenced and not 100% correct.