Question

based on a poll, 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? the probability that at least one of them is careful about personal information is (round to three decimal places as needed.)

Explanation:

Step1: Find the probability of a user not being careful

The probability that an internet - user is more careful is $p = 0.67$. So the probability that a user is not careful is $q=1 - p=1 - 0.67 = 0.33$.

Step2: Find the probability that none of the four users are careful

Since the selections are independent, the probability that none of the $n = 4$ randomly - selected users are careful is given by the binomial probability formula $P(X = k)=C(n,k)\times p^{k}\times q^{n - k}$, where $n = 4$, $k = 0$, $p = 0.67$, and $q = 0.33$. Here, $C(n,k)=\frac{n!}{k!(n - k)!}$, so $C(4,0)=\frac{4!}{0!(4 - 0)!}=1$. Then $P(X = 0)=C(4,0)\times(0.67)^{0}\times(0.33)^{4}=1\times1\times0.01185921 = 0.01185921$.

Step3: Find the probability that at least one user is careful

The probability that at least one user is careful is the complement of the event that none of the users are careful. Let $A$ be the event that at least one user is careful. Then $P(A)=1 - P(X = 0)$. So $P(A)=1-0.01185921 = 0.98814079\approx0.988$.
The fact that the survey subjects volunteered to respond means that there is a voluntary - response bias. People with strong opinions about personal information and public Wi - Fi are more likely to respond. This may not accurately represent the entire population of internet users, so the result may not be valid for all internet users.

Answer:

$0.988$