a survey finds that 48% of people identify themselves as fans of professional football, 12% as fans of car…

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. $p(f|c)=0.75$ $p(c|f)=0.25$ $p(ccap f)=0.09$ $p(ccap f)=p(fcap c)$ $p(c|f)=p(f|c)$

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. $p(f|c)=0.75$ $p(c|f)=0.25$ $p(ccap f)=0.09$ $p(ccap f)=p(fcap c)$ $p(c|f)=p(f|c)$

Answer

Explanation:

Step1: Recall probability - formula definitions

The formula for conditional probability is $P(A|B)=\frac{P(A\cap B)}{P(B)}$, and the intersection of two events $A$ and $B$ is commutative, i.e., $P(A\cap B)=P(B\cap A)$. Given $P(F) = 0.48$, $P(C)=0.12$, and $P(F\cap C)=0.09$.

Step2: Calculate $P(F|C)$

Using the formula $P(F|C)=\frac{P(F\cap C)}{P(C)}$, substitute $P(F\cap C) = 0.09$ and $P(C)=0.12$. Then $P(F|C)=\frac{0.09}{0.12}=0.75$.

Step3: Calculate $P(C|F)$

Using the formula $P(C|F)=\frac{P(C\cap F)}{P(F)}$, substitute $P(C\cap F) = 0.09$ and $P(F)=0.48$. Then $P(C|F)=\frac{0.09}{0.48}=\frac{3}{16}= 0.1875$.

Step4: Analyze the properties of intersection

By the commutative - property of the intersection of two events, $P(C\cap F)=P(F\cap C)$. Since $P(C\cap F) = 0.09$ and $P(F\cap C)=0.09$, this property holds.

Answer:

$P(F|C) = 0.75$, $P(C\cap F)=0.09$, $P(C\cap F)=P(F\cap C)$