Two persons $A$ and $B$ take part in a shooting competition. $A$ can hit the target with a probability of $0.6$. $B$ can hit the target with a probability of $0.8$. $A$ takes the first shot,after which they shoot alternately. The probability that $A$ wins the competition is

If $A$ and $B$ are independent events and $P(A) = p, P(B) = 2p$,and $P(\text{exactly one from } A \text{ and } B) = \frac{5}{9}$,then find the value of $p$.

$A$ bag contains $20$ coins. If the probability that the bag contains exactly $4$ biased coins is $1/3$ and the probability that it contains exactly $5$ biased coins is $2/3$,then the probability that all the biased coins are sorted out from the bag in exactly $10$ draws is:

The probabilities that players $A$ and $B$ of a team are selected for the captaincy for a tournament are $0.6$ and $0.4$, respectively. If $A$ is selected as the captain, the probability that the team wins the tournament is $0.8$ and if $B$ is selected as the captain, the probability that the team wins the tournament is $0.7$. Then the probability, that the team wins the tournament, is:

Two players,$P_1$ and $P_2$,play a game against each other. In every round,each player rolls a fair die once. Let $x$ and $y$ denote the outcomes for $P_1$ and $P_2$. If $x > y$,$P_1$ scores $5$ points and $P_2$ scores $0$. If $x = y$,each scores $2$ points. If $x < y$,$P_1$ scores $0$ and $P_2$ scores $5$. Let $X_n$ and $Y_n$ be the total scores of $P_1$ and $P_2$ after $n$ rounds. Match the following:

List-$I$	List-$II$
$(I)$ Probability of $(X_2 \geq Y_2)$ is	$(P)$ $\frac{3}{8}$
$(II)$ Probability of $(X_2 > Y_2)$ is	$(Q)$ $\frac{11}{16}$
$(III)$ Probability of $(X_3 = Y_3)$ is	$(R)$ $\frac{5}{16}$
$(IV)$ Probability of $(X_3 > Y_3)$ is	$(S)$ $\frac{355}{864}$
	$(T)$ $\frac{77}{432}$