Question

given: p(e)=0.35, p(f)=0.53, and p(e∪f)=0.71. find p(e∩f) and p(f|e). a) p(e∩f)=0.19 and p(f|e)=0.49 b) p(e∩f)=0.29 and p(f|e)=0.32 c) p(e∩f)=0.17 and p(f|e)=0.49 d) p(e∩f)=0.19 and p(f|e)=0.53 e) p(e∩f)=0.17 and p(f|e)=0.53

Explanation:

Step1: Recall the formula for conditional probability

The formula for conditional probability is $P(F|E)=\frac{P(E\cap F)}{P(E)}$. We can re - arrange it to find $P(E\cap F)=P(F|E)\times P(E)$. Also, recall the addition rule for probability $P(E\cup F)=P(E)+P(F)-P(E\cap F)$.

Step2: First, find $P(E\cap F)$ using the addition rule

We know that $P(E\cup F) = P(E)+P(F)-P(E\cap F)$. Rearranging for $P(E\cap F)$ gives $P(E\cap F)=P(E)+P(F)-P(E\cup F)$. Substitute $P(E) = 0.35$, $P(F)=0.53$ and $P(E\cup F)=0.71$ into the formula.
$P(E\cap F)=0.35 + 0.53-0.71=0.17$.

Step3: Then, find $P(F|E)$ using the conditional - probability formula

$P(F|E)=\frac{P(E\cap F)}{P(E)}$. Substitute $P(E\cap F) = 0.17$ and $P(E)=0.35$ into the formula.
$P(F|E)=\frac{0.17}{0.35}\approx0.49$.

Answer:

C. $P(E\cap F)=0.17$ and $P(F|E)=0.49$