Question

Suppose p(a) = 0.40 and p(a 

Answer 1

Between the probability of union and intersection, it's not clear what you're supposed to compute. (I would guess it's the probability of union.) But we do know that

$P(A\cup B)+P(A\cap B)=P(A)+P(B)$


For parts (a) and (b), you're given everything you need to determine $P(B)$.

For part (c), if $A$ and $B$ are mutually exclusive, then $P(A\cap B)=0$, so $P(A\cup B)=P(A)+P(B)$. If the given probability is $P(A\cup B)=0.55$, then you can find $P(B)=0.15$. But if this given probability is for the intersection, finding $P(B)$ is impossible.


For part (d), if $A$ and $B$ are independent, then $P(A\cap B)=P(A)\cdot P(B)$.