The probability that a man will be alive in $20$ years is $\frac{3}{5}$ and the probability that his wife will be alive in $20$ years is $\frac{2}{3}$. Then the probability that at least one will be alive in $20$ years, 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 $P \left( B \cap A ^{\prime}\right)$.

India plays two matches each with West Indies and Australia. In any match the probabilities of India getting point $0, 1$ and $2$ are $0.45, 0.05$ and $0.50$ respectively. Assuming that the outcomes are independents, the probability of India getting at least $7$ points is

Let $S=\{1,2,3, \ldots, 2022\}$. Then the probability, that a randomly chosen number $n$ from the set $S$ such that $\operatorname{HCF}( n , 2022)=1$, is.

Let $A$ and $B$ be events for which $P(A) = x$, $P(B) = y,$$P(A \cap B) = z,$ then $P(\bar A \cap B)$ equals

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