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?

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 $1$.

If $E$ and $F$ are independent events such that $0 < P(E) < 1$ and $0 < P(F) < 1,$ then

For three events $A$, $B$, and $C$ of a sample space, $P(\text{exactly one of } A \text{ or } B \text{ occurs}) = P(\text{exactly one of } B \text{ or } C \text{ occurs}) = P(\text{exactly one of } C \text{ or } A \text{ occurs}) = \frac{1}{4}$. If the probability of all the three events occurring simultaneously is $\frac{1}{16}$, then the probability that at least one of the events occurs is:

Consider the following statements:
Assertion $(A)$: If $P_1, P_2, P_3$ are probabilities of occurrence of three independent events, then the probability of occurrence of at least one of them is $1 - [(1 - P_1)(1 - P_2)(1 - P_3)]$.
Reason $(R)$: For any three independent events $A, B$, and $C$, $P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A)P(B) - P(A)P(C) - P(B)P(C) + P(A)P(B)P(C)$.
The correct option among the following is:

Two balls are drawn at random with replacement from a box containing $10$ black and $8$ red balls. Find the probability that one of them is black and the other is red.