When a coin is tossed $6$ times,the probability of getting more heads than tails is

In a Binomial distribution,if $n$ is the number of trials and the mean and variance are $4$ and $3$ respectively,then $2^{32} P\left(X=\frac{n}{2}\right)=$

One out of $9$ ships is likely to sink when they are set on sail. When $6$ ships are set on sail,the probability that exactly $3$ of them will not arrive safely is

$A$ box contains $15$ green and $10$ yellow balls. If $10$ balls are randomly drawn,one by one,with replacement,then the variance of the number of green balls drawn 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 $\frac{\mu \sigma^2}{P(X=3)}=$

$A$ fair coin is tossed a fixed number of times. If the probability of getting exactly $3$ heads equals the probability of getting exactly $5$ heads, then the probability of getting exactly one head is