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:

Consider an experiment of tossing a coin repeatedly until the outcomes of two consecutive tosses are the same. If the probability of a random toss resulting in a head is $\frac{1}{3}$,then the probability that the experiment stops with heads is:

For the three events $A, B$ and $C$,$P$ (exactly one of the events $A$ or $B$ occurs) = $P$ (exactly one of the events $B$ or $C$ occurs) = $P$ (exactly one of the events $C$ or $A$ occurs) = $p$ and $P$ (all the three events occur simultaneously) = $p^2$,where $0 < p < 1/2$. Then the probability of at least one of the three events $A, B$ and $C$ occurring is

$A$ coin is tossed $m + n$ times,where $m \ge n.$ The probability of getting at least $m$ consecutive heads is

Let $X_n = \{1, 2, 3, \ldots, n\}$ and let a subset $A$ of $X_n$ be chosen such that every pair of elements of $A$ differ by at least $3$. (For example,if $n = 5$,$A$ can be $\phi, \{2\}$ or $\{1, 5\}$ among others). When $n = 10$,let the probability that $1 \in A$ be $p$ and let the probability that $2 \in A$ be $q$. Then,

If the letters of the word $REGULATIONS$ are arranged in such a way that the relative positions of the letters of the word $GULATIONS$ remain the same,then the probability that there are exactly $4$ letters between $R$ and $E$ is