There are $3$ bags $A, B$ & $C$. Bag $A$ contains $1$ Red & $2$ Green balls,bag $B$ contains $2$ Red & $1$ Green balls and bag $C$ contains only $1$ Green ball. One ball is drawn from bag $A$ & put into bag $B$,then one ball is drawn from $B$ & put into bag $C$,& finally one ball is drawn from bag $C$ & put into bag $A$. When this operation is completed,what is the probability that bag $A$ contains $2$ Red & $1$ Green balls?

Two coins are tossed. Let $A$ be the event that the first coin shows a head and $B$ be the event that the second coin shows a tail. The two events $A$ and $B$ are:

$A$ ship is fitted with three engines $E_1, E_2$,and $E_3$. The engines function independently of each other with respective probabilities $\frac{1}{2}, \frac{1}{4}$,and $\frac{1}{4}$. For the ship to be operational,at least two of its engines must function. Let $X$ denote the event that the ship is operational and let $X_1, X_2$,and $X_3$ denote respectively the events that the engines $E_1, E_2$,and $E_3$ are functioning. Which of the following is (are) true?
$(A) P(X_1^c \mid X) = \frac{3}{16}$
$(B) P(\text{Exactly two engines are functioning} \mid X) = \frac{7}{8}$
$(C) P(X \mid X_2) = \frac{5}{16}$
$(D) P(X \mid X_1) = \frac{7}{16}$

The corners of regular tetrahedrons are numbered $1, 2, 3, 4$. Three tetrahedrons are tossed. The probability that the sum of the upward corners will be $5$ is

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

Consider the system of equations $ax+by=0, cx+dy=0$,where $a, b, c, d \in \{0, 1\}$.
$STATEMENT-1$: The probability that the system of equations has a unique solution is $3/8$.
$STATEMENT-2$: The probability that the system of equations has a solution is $1$.