Two persons $A$ and $B$ are throwing an unbiased six-faced dice alternatively,with the condition that the person who throws $3$ first wins the game. If $A$ starts the game,then the probabilities of $A$ and $B$ winning the game are,respectively:

Let $E$ and $F$ be two independent events. The probability that both $E$ and $F$ happen is $\frac{1}{12}$ and the probability that neither $E$ nor $F$ happens is $\frac{1}{2}$. Then a value of $\frac{P(E)}{P(F)}$ 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:

Let $S$ be the sample space of a random experiment and $P$ be a probability function defined on the power set of $S$. Two events $A$ and $B$ of the random experiment are called independent if

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 $IIT$?

$A$ and $B$ are two independent events. $P(A)=\frac{2}{5}, P(B)=\frac{1}{3}$. Match the following List-$I$ with List-$II$.

List-$I$	List-$II$
$(A) P(\overline{A} \cup B)$	$(I) \frac{2}{3}$
$(B) P(\frac{A}{\overline{B}})$	$(II) \frac{11}{15}$
$(C) P(A \cup B)$	$(III) \frac{3}{5}$