If the numbers appearing on the two throws of a fair six-faced die are $\alpha$ and $\beta$,then the probability that $x^{2}+\alpha x+\beta > 0$ for all $x \in R$ 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-$2$: $P(X) + P(Y) = 2P(X \cap Y).$

$A$ and $B$ are independent events with $P(A)=\frac{1}{4}$ and $P(A \cup B)=2 P(B)-P(A)$,then $P(B)$ is

For the three events $A, B$ and $C$,$P$ (exactly one of the events $A$ or $B$ occurs) = $P$ (exactly one of the events $B$ or $C$ occurs) = $P$ (exactly one of the events $C$ or $A$ occurs) = $p$ and $P$ (all the three events occur simultaneously) = $p^2$,where $0 < p < 1/2$. Then the probability of at least one of the three events $A, B$ and $C$ occurring is

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:

Two numbers are selected randomly from the set $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ without replacement one by one. The probability that the minimum of the two numbers is divisible by $3$ or the maximum of the two numbers is divisible by $4$ is: