$A$ fair die is thrown until $2$ appears. Then the probability that $2$ appears in an even number of throws is

$A$ fair die is rolled indefinitely. Player $A$ wins if two consecutive rolls show $3$ or $5$, and player $B$ wins if two consecutive rolls show $1$ or $2$ or $4$ or $6$. The probability that player $A$ wins is:

Three urns respectively contain $2$ white and $3$ black,$3$ white and $2$ black,and $1$ white and $4$ black balls. If one ball is drawn from each urn,then the probability that the selection contains $1$ black and $2$ white balls 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	$(S)$ $\frac{355}{864}$
	$(T)$ $\frac{77}{432}$

$2$ aeroplanes $I$ and $II$ bomb a target in succession. The probabilities of $I$ and $II$ scoring a hit correctly are $0.3$ and $0.2$ respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the $2^{nd}$ plane is

$S$ is the sample space and $A, B$ are two events of a random experiment. Match the items of List-$A$ with the items of List-$B$.

List-$A$	List-$B$
$(I)$ $A, B$ are mutually exclusive events	$(i)$ $P(A \cap B) = P(B) - P(\bar{A})$
$(II)$ $A, B$ are independent events	$(ii)$ $P(A) \leq P(B)$
$(III)$ $A \cap B = A$	$(iii)$ $P(\frac{\bar{A}}{B}) = 1 - P(A)$
$(IV)$ $A \cup B = S$	$(iv)$ $P(A \cup B) = P(A) + P(B)$
	$(v)$ $P(A) + P(B) = 2$