If the mean and variance of a binomial distribution are $4$ and $2$ respectively,then the probability of $2$ successes of that binomial variate $X$ is:

Find the mean of the Binomial distribution $B\left(4, \frac{1}{3}\right)$.

For a binomial variate $X \sim B(n, p)$,the difference between the mean and variance is $1$ and the difference between their squares is $11$. If the probability $P(X=2) = m\left(\frac{5}{6}\right)^n$ and $n=36$,then $m : n =$

The mean and variance of a binomial distribution are $6$ and $4$. The parameter $n$ is

If $x$ denotes the number of sixes in four consecutive throws of a dice,then $P(x = 4)$ is

In a tournament,a team plays $10$ matches with probabilities of winning and losing each match as $\frac{1}{3}$ and $\frac{2}{3}$ respectively. Let $x$ be the number of matches that the team wins,and $y$ be the number of matches that the team loses. If the probability $P(|x-y| \leq 2)$ is $p$,then $3^9 p$ equals....................