Question

a survey finds that 48% of people identify themselves as fans of professional football, 12% as fans of car racing, and 9% as fans of both professional football and car racing. let event f be choosing a person who is a fan of professional football and let event c be choosing a person who is a fan of car racing. which statements are true? select three options. \\( p(f|c) = 0.75 \\) \\( p(c|f) = 0.25 \\) \\( p(c cap f) = 0.09 \\) \\( p(c cap f) = p(f cap c) \\) \\( p(c|f) = p(f|c) \\)

Explanation:

Step1: Recall Conditional Probability Formula

The formula for conditional probability is \( P(A|B) = \frac{P(A \cap B)}{P(B)} \), and also \( P(A \cap B) = P(B \cap A) \) (intersection is commutative). We know \( P(F) = 0.48 \), \( P(C) = 0.12 \), \( P(F \cap C) = 0.09 \).

Step2: Check \( P(F|C) \)

Using the formula, \( P(F|C) = \frac{P(F \cap C)}{P(C)} = \frac{0.09}{0.12} = 0.75 \). So this statement is true.

Step3: Check \( P(C|F) \)

Using the formula, \( P(C|F) = \frac{P(C \cap F)}{P(F)} = \frac{0.09}{0.48} = 0.1875
eq 0.25 \). So this statement is false.

Step4: Check \( P(C \cap F) \)

We are given that 9% are fans of both, so \( P(C \cap F) = 0.09 \). This statement is true.

Step5: Check \( P(C \cap F) = P(F \cap C) \)

By the commutative property of intersection, \( A \cap B = B \cap A \), so their probabilities are equal. This statement is true.

Step6: Check \( P(C|F) = P(F|C) \)

We calculated \( P(C|F) = 0.1875 \) and \( P(F|C) = 0.75 \), which are not equal. So this statement is false.

Answer:

• \( P(F|C) = 0.75 \)
• \( P(C \cap F) = 0.09 \)
• \( P(C \cap F) = P(F \cap C) \)