The probability of a man hitting a target is $\frac{2}{5}$. He fires at the target $k$ times ($k$ is a given number). The minimum value of $k$ such that the probability of hitting the target at least once is more than $\frac{7}{10}$ is:

$A$ 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

$A$ random variable $X \sim B(n, p)$ follows a binomial distribution with $n = 6$. If $9P(X = 4) = P(X = 2)$, then the probability of success $p$ is:

Numbers are selected at random,one at a time from the two-digit numbers $00, 01, 02, \dots, 99$ with replacement. An event $E$ occurs only if the product of the two digits of a selected number is $24$. If four numbers are selected,then the probability that the event $E$ occurs at least $3$ times is:

If the mean and the variance of a binomial variate $X$ are $1$ and $0.75$ respectively, then which of the following is true?

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