To solve this problem, we need to determine the conditional probability that a guest also ate a turkey leg given that they ate a coney dog. In probability terms, we are looking for [tex]\( P(T \mid C) \)[/tex], which is the probability of [tex]\( T \)[/tex] (eating a turkey leg) given [tex]\( C \)[/tex] (eating a coney dog).
We are provided with the following probabilities:
- The probability of a guest eating both a coney dog and a turkey leg ([tex]\( P(C \text{ and } T) \)[/tex]) is [tex]\( \frac{3}{16} \)[/tex].
- The probability of a guest eating a coney dog ([tex]\( P(C) \)[/tex]) is [tex]\( \frac{7}{17} \)[/tex].
The formula for conditional probability is given by:
[tex]\[ P(T \mid C) = \frac{P(C \text{ and } T)}{P(C)} \][/tex]
Let's plug the given probabilities into the formula:
[tex]\[ P(T \mid C) = \frac{\frac{3}{16}}{\frac{7}{17}} \][/tex]
To divide these fractions, multiply by the reciprocal of the denominator:
[tex]\[ P(T \mid C) = \frac{3}{16} \times \frac{17}{7} \][/tex]
Multiply the numerators and the denominators:
[tex]\[ P(T \mid C) = \frac{3 \times 17}{16 \times 7} \][/tex]
[tex]\[ P(T \mid C) = \frac{51}{112} \][/tex]
Therefore, the probability that a guest who had a coney dog also had a turkey leg is [tex]\( \frac{51}{112} \)[/tex].
The correct answer is:
[tex]\[ \frac{51}{112} \][/tex]
