Let $X$ and $Y$ be two events such that $P(X \cup Y) = P(X \cap Y)$.
Statement $1$: $P(X \cap Y') = P(X' \cap Y) = 0$.
Statement $2$: $P(X) + P(Y) = 2P(X \cap Y)$.

Three persons $P, Q$ and $R$ independently try to hit a target. If the probabilities of their hitting the target are $\frac{3}{4}, \frac{1}{2}$ and $\frac{5}{8}$ respectively,then the probability that the target is hit by $P$ or $Q$ but not by $R$,is

$A$ and $B$ are independent events. If $P(A \cup B)=0.5$ and $P(A)=0.2$,then $P(B) = $ . . . . . . . (in $/8$)

Cards are drawn one by one without replacement from a pack of $52$ cards. The probability that $10$ cards will precede the first ace is

Let there be three independent events $E_{1}, E_{2}$ and $E_{3}$. The probability that only $E_{1}$ occurs is $\alpha$,only $E_{2}$ occurs is $\beta$ and only $E_{3}$ occurs is $\gamma$. Let $p$ denote the probability that none of the events occur,which satisfies the equations $(\alpha - 2\beta)p = \alpha\beta$ and $(\beta - 3\gamma)p = 2\beta\gamma$. All the given probabilities are assumed to lie in the interval $(0, 1)$. Then,$\frac{\text{Probability of occurrence of } E_{1}}{\text{Probability of occurrence of } E_{3}}$ is equal to ..........

In a game, two dice are thrown simultaneously by a person $A$ and two cards are drawn at random simultaneously from a pack of $52$ playing cards by a person $B$. They win the game if $A$ gets a prime score as the sum of the numbers appearing on both the dice and $B$ gets a face card and a card having a prime number. Then the probability that both $A$ and $B$ win is: