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∩f) = 0.09 p(c∩f) = p(f∩c) p(c|f) = p(f|c)

Explanation:

Step1: Recall the formula for conditional probability

The formula for conditional probability is \(P(A|B)=\frac{P(A\cap B)}{P(B)}\).

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

Given \(P(F) = 0.48\), \(P(C)=0.12\), \(P(F\cap C)=0.09\).
Using the formula \(P(F|C)=\frac{P(F\cap C)}{P(C)}\), substitute the values: \(P(F|C)=\frac{0.09}{0.12}= 0.75\).

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

Using the formula \(P(C|F)=\frac{P(C\cap F)}{P(F)}\), substitute the values: \(P(C|F)=\frac{0.09}{0.48}=0.1875
eq0.25\).

Step4: Analyze \(P(C\cap F)\) and \(P(F\cap C)\)

By the commutative property of intersection in probability, \(P(C\cap F)=P(F\cap C)\). And \(P(C\cap F) = 0.09\) (given as the probability of being a fan of both).

Step5: Check \(P(C|F)\) and \(P(F|C)\) equality

Since \(P(C|F)=\frac{P(C\cap F)}{P(F)}\) and \(P(F|C)=\frac{P(F\cap C)}{P(C)}\), and \(P(F)
eq P(C)\), \(P(C|F)
eq P(F|C)\)

Answer:

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