Question

probability and statistics
a survey finds that 45% of people identify themselves as fans of professional football, 25% 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? choose three correct answers.
p(c|f)=p(f|c)
p(c∩f)=p(f∩c)
p(c|f)=0.25
p(c∩f)=0.09
p(f|c)=0.75

Explanation:

Step1: Analyze \(P(C\cap F)\) and \(P(F\cap C)\)

By the commutative property of intersection in probability, \(P(C\cap F)=P(F\cap C)\).

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

Given \(P(F) = 0.36\) (36% of people are fans of professional football), \(P(C)=0.25\) (25% of people are fans of car - racing) and the formula \(P(C\cap F)=P(C|F)P(F)=P(F|C)P(C)\). Since \(P(C\cap F)\) (percentage of people who are fans of both) is \(9\%\) or \(0.09\) (because intersection probability is calculated as \(P(C\cap F)\)).

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

Using the formula \(P(C|F)=\frac{P(C\cap F)}{P(F)}\), substituting \(P(C\cap F) = 0.09\) and \(P(F)=0.36\), we get \(P(C|F)=\frac{0.09}{0.36}=0.25\).

Step4: Calculate \(P(F|C)\)

Using the formula \(P(F|C)=\frac{P(F\cap C)}{P(C)}\), substituting \(P(F\cap C) = 0.09\) and \(P(C)=0.25\), we get \(P(F|C)=\frac{0.09}{0.25}=0.36
eq0.75\). Also, in general \(P(C|F)=\frac{P(C\cap F)}{P(F)}\) and \(P(F|C)=\frac{P(F\cap C)}{P(C)}\), and since \(P(F)
eq P(C)\) (unless \(P(F) = P(C)\) which is not the case here), \(P(C|F)
eq P(F|C)\)

Answer:

\(P(C\cap F)=P(F\cap C)\), \(P(C|F) = 0.25\), \(P(C\cap F)=0.09\)