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?

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

For two events $A$ and $B$,if $P(A) = P(A|B) = \frac{1}{4}$ and $P(B|A) = \frac{1}{2}$,then which of the following is true?

For three events $A, B$ and $C$,$P(\text{Exactly one of } A \text{ or } B \text{ occurs}) = P(\text{Exactly one of } B \text{ or } C \text{ occurs}) = P(\text{Exactly one of } C \text{ or } A \text{ occurs}) = \frac{1}{4}$ and $P(\text{All the three events occur simultaneously}) = \frac{1}{16}$. Then the probability that at least one of the events occurs is:

Three dice are thrown simultaneously and the sum of the numbers appeared on them is noted. If $A$ is the event of getting a sum greater than $14$ and $B$ is the event of getting a sum which is a multiple of $3$,then $P(A \cap \overline{B}) + P(\overline{A} \cap B) = $

Two players $A$ and $B$ are alternately throwing a coin and a die together. $A$ player who first throws a head and a $6$ wins the game. If $A$ starts the game,then the probability that $B$ wins the game is: