Given two independent events,if the probability that exactly one of them occurs is $\frac{26}{49}$ and the probability that none of them occurs is $\frac{15}{49}$,then the probability of the more probable of the two events is (in $/7$)

There are three bags $B_1$,$B_2$,and $B_3$ containing $2$ Red and $3$ White,$5$ Red and $5$ White,and $3$ Red and $2$ White balls respectively. $A$ ball is drawn from bag $B_1$ and placed in bag $B_2$,then a ball is drawn from bag $B_2$ and placed in bag $B_3$,then a ball is drawn from bag $B_3$. The number of ways in which this process can be completed,if the same colour balls are used in the first and second transfers (assume all balls to be distinct),is

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 $A$ denote the set of all $4$-digit natural numbers with no digit being $0$. Let $B \subset A$ consist of all numbers $x$ such that no permutation of the digits of $x$ gives a number that is divisible by $4$. Then the probability of drawing a number from $B$ with all even digits is

$A$ coin is tossed $m + n$ times,where $m \ge n.$ The probability of getting at least $m$ consecutive heads is

$A$ bag contains $2$ white,$3$ green,and $5$ red balls. If three balls are drawn one after the other without replacement,then the probability that the last ball drawn was red is