I'd love if someone could go over how to do this problem.
Given $P(A\cap B)=0.4$, $P(A\cap B^c)=0.2$, $P(A^c\cap B)=0.3$.
Determine:
$P(A)$:
$P(B):$
$P(A|B):$
$P(B|A):$
I'd love if someone could go over how to do this problem.
Given $P(A\cap B)=0.4$, $P(A\cap B^c)=0.2$, $P(A^c\cap B)=0.3$.
Determine:
$P(A)$:
$P(B):$
$P(A|B):$
$P(B|A):$
$(A\cap B)\cup(A\cap B') = ((A\cap B)\cup A) \cap ((A\cap B)\cup(B')) \\ =A$
$P(A) = P((A\cap B)\cup(A\cap B')) = P(A\cap B) + P(A\cap B) = 0.6$
Since $(A\cap B)$ and $(A\cap B')$ are disjoint. You can do $P(B)$ in a similar manner.
$P(B|A) = \frac{P(B\cap A)}{P(A)} = \frac{0.4}{0.6}$
$P(A|B) = \frac{P(A\cap B)}{P(B)}$ which can be solved when you get $P(B)$
You need three things;
Study and apply these three important theorems and you should be able to solve your problem.
NB: It is advisable to read study the content of the three links.