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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 [tex]\(\frac{3}{16}\)[/tex]. The probability of eating a coney dog is [tex]\(\frac{7}{17}\)[/tex]. Given that a guest ate a coney dog, what is the probability the guest also ate a turkey leg?

A. [tex]\(\frac{51}{112}\)[/tex]

B. [tex]\(\frac{3}{112}\)[/tex]

C. [tex]\(\frac{21}{272}\)[/tex]

D. [tex]\(\frac{7}{272}\)[/tex]



Answer :

GinnyAnswer
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]

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