Question

question number 2. given: p(e)=0.37, p(f)=0.53, and p(e∪f)=0.72. find p(e∩f). o 0.82 o 0.28 o 0.80 o 0.18 o 0.20

Explanation:

Step1: Recall the formula

Use the formula $P(E\cup F)=P(E)+P(F)-P(E\cap F)$.

Step2: Rearrange the formula

We can rewrite it as $P(E\cap F)=P(E)+P(F)-P(E\cup F)$.

Step3: Substitute values

Given $P(E) = 0.37$, $P(F)=0.55$ and $P(E\cup F)=0.72$. Then $P(E\cap F)=0.37 + 0.55- 0.72$.

Step4: Calculate

$P(E\cap F)=0.20$.

Answer:

$0.20$