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

Answer

Explanation:

Step1: Recall conditional - probability formula

The formula for conditional probability is $P(A|B)=\frac{P(A\cap B)}{P(B)}$. Also, by the commutative property of intersection, $P(A\cap B) = P(B\cap A)$. 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 statements

For $P(F|C) = 0.75$, since we calculated $P(F|C)=\frac{0.09}{0.12}=0.75$, this statement is true.
For $P(C|F)=0.25$, but we calculated $P(C|F)=\frac{0.09}{0.48}=0.1875\neq0.25$, this statement is false.
For $P(C\cap F)=0.09$, given in the problem, this statement is true.
For $P(C\cap F)=P(F\cap C)$, by the commutative property of intersection, this statement is true.
For $P(C|F)=P(F|C)$, since $P(C|F)=0.1875$ and $P(F|C)=0.75$, this statement is false.

Answer:

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