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:

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:

Let $X$ and $Y$ be two events such that $P(X \mid Y)=\frac{1}{2}$,$P(Y \mid X)=\frac{1}{3}$,and $P(X \cap Y)=\frac{1}{6}$. Which of the following is (are) correct?
$(A)$ $P(X \cup Y)=\frac{2}{3}$
$(B)$ $X$ and $Y$ are independent
$(C)$ $X$ and $Y$ are not independent
$(D)$ $P(X^C \cap Y)=\frac{1}{3}$

You are given a box containing $20$ cards. Out of these,$10$ cards have the letter $I$ printed on them,and the other $10$ cards have the letter $T$ printed on them. If you draw three cards one after another with replacement,what is the probability of forming the word $IIT$?

Two persons $A$ and $B$ throw a fair die (six-faced cube with faces numbered from $1$ to $6$) alternately,starting with $A$. The first person to get an outcome different from the previous one thrown by the opponent wins. The probability that $B$ wins 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.