If $P(B) = \frac{3}{4}$, $P(A \cap B \cap \bar C) = \frac{1}{3}{\rm{ }}$ and $P(\bar A \cap B \cap \bar C) = \frac{1}{3},$ then $P(B \cap C)$ is
$\frac{1}{{12}}$
$\frac{1}{6}$
$\frac{1}{{15}}$
$\frac{1}{9}$
If $P\,(A) = \frac{1}{4},\,\,P\,(B) = \frac{5}{8}$ and $P\,(A \cup B) = \frac{3}{4},$ then $P\,(A \cap B) = $
If $A$ and $B$ are two events such that $P(A) = \frac{1}{2}$ and $P(B) = \frac{2}{3},$ then
Let $A$ and $B$ be independent events with $P(A)=0.3$ and $P(B)=0.4$. Find $P(A \cup B)$
Four persons can hit a target correctly with probabilities $\frac{1}{2},\frac{1}{3},\frac{1}{4}$ and $\frac {1}{8}$ respectively. If all hit at the target independently, then the probability that the target would be hit, is
If $A$ and $B$ are two events of a random experiment, $P\,(A) = 0.25$, $P\,(B) = 0.5$ and $P\,(A \cap B) = 0.15,$ then $P\,(A \cap \bar B) = $