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 conditional - probability formula

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

Step2: Identify given probabilities

We are given that $P(F) = 0.48$, $P(C)=0.12$, and $P(F\cap C)=0.09$.

Step3: Calculate $P(F|C)$

$P(F|C)=\frac{P(F\cap C)}{P(C)}=\frac{0.09}{0.12}= 0.75$.

Step4: Calculate $P(C|F)$

$P(C|F)=\frac{P(C\cap F)}{P(F)}=\frac{0.09}{0.48}=0.1875
eq0.25$.

Step5: Commutative property of intersection

By the commutative property of intersection of events, $P(C\cap F)=P(F\cap C)$. Since $P(F\cap C) = 0.09$, $P(C\cap F)=0.09$. And in general, for any two events $A$ and $B$, $P(A\cap B)=P(B\cap A)$. Also, $P(C|F)=\frac{P(C\cap F)}{P(F)}$ and $P(F|C)=\frac{P(F\cap C)}{P(C)}$, and $P(C|F)
eq P(F|C)$ in general (except in some special - cases which is not the case here as calculated above).

Answer:

A. $P(F|C)=0.75$
C. $P(C\cap F)=0.09$
D. $P(C\cap F)=P(F\cap C)$