What is the probability of getting an even number exactly $3$ times when a die is thrown $5$ times?

In tossing $10$ coins,the probability of getting exactly $5$ heads is

The mean and variance of a binomial distribution are $4$ and $3$ respectively. Then,the probability of getting exactly six successes in this distribution is:

An irregular six-faced die is thrown. The probability that in $5$ throws it will give $3$ even numbers is twice the probability that it will give $2$ even numbers. The number of times,in $6804$ sets of $5$ throws,you expect to get no even number is:

$A$ variable $X$ takes values $0, 0, 2, 6, 12, 20, ..., n(n-1)$ with frequencies $^nC_0, ^nC_1, ^nC_2, ^nC_3, ^nC_4, ^nC_5, ..., ^nC_n$,respectively. If the mean of this data is $60$,then its median is:

$7$ coins are tossed simultaneously and the number of heads turned up is denoted by the random variable $X$. If $\mu$ is the mean and $\sigma^2$ is the variance of $X$,then $\frac{\mu \sigma^2}{P(X=3)}=$