The probability that the product of the outcomes when three dice are rolled simultaneously is divisible by $4$ is equal to

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	$(S)$ $\frac{355}{864}$
	$(T)$ $\frac{77}{432}$

Three persons $P, Q$ and $R$ independently try to hit a target. If the probabilities of their hitting the target are $\frac{3}{4}, \frac{1}{2}$ and $\frac{5}{8}$ respectively,then the probability that the target is hit by $P$ or $Q$ but not by $R$,is

Let a computer program generate only the digits $0$ and $1$ to form a string of binary numbers. The probability of occurrence of $0$ at even places is $\frac{1}{2}$ and the probability of occurrence of $0$ at odd places is $\frac{1}{3}$. Then the probability that $'10'$ is followed by $'01'$ is equal to:

Let $\omega$ be a complex cube root of unity with $\omega \neq 1$. $A$ fair die is thrown three times. If $r_1, r_2$ and $r_3$ are the numbers obtained on the die,then the probability that $\omega^{r_1}+\omega^{r_2}+\omega^{r_3}=0$ is

Cards are drawn one by one without replacement from a pack of $52$ cards. The probability that $10$ cards will precede the first ace is