The probability that a person who undergoes a certain operation will survive is $0.2$. If $5$ patients undergo similar operations,then the probability that exactly four will survive is

Let $\lim _{t \rightarrow 0}(1+5 t)^{\frac{1}{t}}=K$ and $X$ be the random variable representing the number of successes in $100$ independent trials. If the probability of success in each trial is $0.05$, then the probability of getting at least one success is

$A$ random variable $X$ follows the binomial distribution and $X \sim B(n, 0.3)$. If the mean of $X$ is three times as large as the standard deviation of $X$,then $n=$

If the sum of the mean and the variance of a binomial distribution for $5$ trials is $1 \cdot 8$,then $p=$

For a binomial distribution,the mean and variance are $6$ and $4$ respectively. Find the value of $n$.

In a binomial distribution,the mean is $15$ and the variance is $10$. Then the parameter $n$ is: