If $n$ positive integers are taken at random and multiplied together,the probability that the last digit of the product is $2, 4, 6$ or $8$ is:

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

All face cards from a pack of $52$ playing cards are removed. From the remaining $40$ cards,two are drawn randomly without replacement. The probability of drawing a pair (cards of the same denomination) is:

Let $|X|$ denote the number of elements in set $X$. Let $S = \{1, 2, 3, 4, 5, 6\}$ be a sample space,where each element is equally likely to occur. If $A$ and $B$ are independent events associated with $S$,then the number of ordered pairs $(A, B)$ such that $1 \leq |B| < |A|$ equals:

$A$ pack of cards contains $4$ aces,$4$ kings,$4$ queens,and $4$ jacks. Two cards are drawn at random. The probability that at least one of these is an ace is:

If $P(A') + P(B') P(A \cup B) = 0.7$,then $P(A') + P(B')$ is