Question

select the correct answer. at a regional fair, the fair staff conducts a survey of the food choices the guests made during their visit. the probability of a guest eating both a coney dog and a turkey leg is 3 out of 16. the probability of eating a coney dog is 7 out of 17. given that a guest ate a coney dog, what is the probability the guest also ate a turkey leg? (\frac{7}{37}) (\frac{1}{17}) (\frac{3}{37}) (\frac{3}{17})

Explanation:

Step1: Recall probability - multiplication rule

For independent events \(A\) and \(B\), the probability of both \(A\) and \(B\) occurring is \(P(A\cap B)=P(A)\times P(B)\). Here, the event of eating a coney - dog and the event of eating a turkey leg are assumed to be independent. Let \(A\) be the event of eating a coney - dog and \(B\) be the event of eating a turkey leg. We are given \(P(A)=\frac{7}{17}\) and \(P(B)=\frac{3}{16}\).

Step2: Calculate the joint probability

\(P(A\cap B)=\frac{7}{17}\times\frac{3}{16}=\frac{21}{272}\)

Answer:

\(\frac{21}{272}\)