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

If three dice are thrown at a time, then the probability of getting the sum of the numbers on them as a prime number is

Two players $A$ and $B$ alternatively toss $3$ coins simultaneously. The player who gets $2$ heads and $1$ tail first, wins the game. If the game continues until someone wins and if $A$ begins the game, the probability that $B$ wins the game is

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 $0 < P(A) < 1$,$0 < P(B) < 1$ and $P(A \cup B) = P(A) + P(B) - P(A)P(B).$ Then

Let $A, B,$ and $C$ be $3$ independent events such that $P(A) = 1/3, P(B) = 1/2,$ and $P(C) = 1/4$. Find the probability that exactly $2$ of the $3$ events occur.