If $P(A)=\frac{6}{11}, P(B)=\frac{5}{11}$ and $P(A \cup B)=\frac{7}{11},$ find $P(A \cap B).$

It is $5:2$ against a husband who is $65$ years old living till he is $85$ and $4:3$ against his wife who is now $58$,living till she is $78$. If the probability that at least one of them will be alive for $20$ years is $k$,then the value of $49k$ is:

$A$ and $B$ are two events such that $P(A)=0.54$,$P(B)=0.69$ and $P(A \cap B)=0.35$. Find $P(A' \cap B')$.

$A, B, C$ are three events,one of which must and only one can happen. The odds in favour of $A$ are $4:6$,and the odds against $B$ are $7:3$. Then,the odds against $C$ are:

Given two independent events $A$ and $B$ such that $P(A) = 0.3$ and $P(B) = 0.6$. Find $P(\text{neither } A \text{ nor } B)$.

The probabilities that $A$ and $B$ speak the truth are $\frac{4}{5}$ and $\frac{3}{4}$ respectively. The probability that they contradict each other when asked to speak on a fact is