Solution

a. Since, 

$P(A \cap B) = P(A)P(B)= 0.12$, the two events are independent.

b. By addition rule,

$$\begin{align*} P(A \cap B)&= P(A) + P(B) - P(A\cup B)\\ &= 0.3 + 0.4 - 0.7\\ &=0 \end{align*}$$

c. If $A$ and $B$ are independent, then $P(A|B)= P(A)=0.3$

d. If $A$ and $B$ are mutually exclusive, then $A \cap B= \emptyset$ and $P(A \cap B)=0$. This implies $P(A|B)= \dfrac{P(A \cap B)}{P(B)}=0$

$Answer$

$a.$Yes

$b.$0 

$c.$0.3

$d.$0

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