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 $/7$)

Functions are formed from the set $A = \{a_1, a_2, a_3\}$ to another set $B = \{b_1, b_2, b_3, b_4, b_5\}$. If a function is selected at random,the probability that it is a one-one function is

Ravi and Rashmi each hold $2$ red cards and $2$ black cards (all four red and all four black cards are identical). Ravi picks a card at random from Rashmi and then Rashmi picks a card at random from Ravi. This process is repeated a second time. Let $p$ be the probability that both have all $4$ cards of the same colour. Then,$p$ satisfies

Four persons $A$,$B$,$C$ and $D$ throw an unbiased die,turn by turn,in succession till one gets an even number and wins the game. What is the probability that $A$ wins if $A$ begins?

Let $X$ and $Y$ be two events such that $P(X \mid Y)=\frac{1}{2}$,$P(Y \mid X)=\frac{1}{3}$,and $P(X \cap Y)=\frac{1}{6}$. Which of the following is (are) correct?
$(A)$ $P(X \cup Y)=\frac{2}{3}$
$(B)$ $X$ and $Y$ are independent
$(C)$ $X$ and $Y$ are not independent
$(D)$ $P(X^C \cap Y)=\frac{1}{3}$

$A$ and $B$ alternately throw a pair of dice. $A$ wins if he throws a sum of $5$ before $B$ throws a sum of $8$,and $B$ wins if he throws a sum of $8$ before $A$ throws a sum of $5$. The probability that $A$ wins,if $A$ makes the first throw,is