The probability that at least one of $A$ and $B$ occurs is $0.6$. If $A$ and $B$ occur simultaneously with probability $0.3$, then $P(A') + P(B') = $
$0.9$
$1.15$
$1.1$
$1.2$
If $P\,(A) = 0.4,\,\,P\,(B) = x,\,\,P\,(A \cup B) = 0.7$ and the events $A$ and $B$ are mutually exclusive, then $x = $
$A , B, C$ try to hit a target simultaneously but independently. Their respective probabilities of hitting targets are $\frac{3}{4},\frac{1}{2},\frac{5}{8}$. The probability that the target is hit by $A$ or $B$ but not by $C$ is
$A$ and $B$ are two independent events. The probability that both $A$ and $B$ occur is $\frac{1}{6}$ and the probability that neither of them occurs is $\frac{1}{3}$. Then the probability of the two events are respectively
$A$ and $B$ are two events such that $P(A)=0.54$, $P(B)=0.69$ and $P(A \cap B)=0.35.$ Find $P \left( A \cap B ^{\prime}\right)$ .
If the odds in favour of an event be $3 : 5$, then the probability of non-occurrence of the event is