Let $X \sim B(n, p)$ with mean $\mu$ and variance $\sigma^2$. If $\mu=2 \sigma^2$ and $\mu+\sigma^2=3$,then $P(X \leq 3)=$

The mean and variance of a binomial distribution are $5$ and $4$ respectively. Then what is $P(X=1)$?

In a box containing $100$ bulbs,$10$ are defective. The probability that out of a sample of $5$ bulbs,none is defective is . . . . . . .

$A$ fair coin is tossed $100$ times. The probability of getting tails an odd number of times is

The discrete random variables $X$ and $Y$ are independent from one another and are defined as $X \sim B(n_1, 0.5)$ and $Y \sim B(n_2, 0.4)$. If the variance of both $X$ and $Y$ is $6$,then $\sqrt{n_1+n_2}=$

$A$ target is to be destroyed in a bombing exercise and there is a $75 \%$ chance that a bomb will hit the target. Assuming that two direct hits are required to destroy the target completely, the minimum number of bombs to be dropped in order that the probability of destroying the target is not less than $99 \%$, is