$A$ die is thrown twice. If getting a number greater than $4$ on the die is considered a success,then the variance of the probability distribution of the number of successes is

$A$ die is thrown five times. If getting an odd number is a success,then the probability of getting at least $4$ successes is

An urn contains $25$ balls of which $10$ balls bear a mark $'X'$ and the remaining $15$ bear a mark $'Y'$. $A$ ball is drawn at random from the urn,its mark is noted down and it is replaced. If $6$ balls are drawn in this way,find the probability that not more than $2$ will bear $'Y'$ mark.

In a Binomial distribution $B(n, p)$,the sum and product of the mean and the variance are $5$ and $6$ respectively,then $6(n+p-q)=$

$A$ man takes a step forward with probability $0.4$ and backward with probability $0.6$. The probability that at the end of eleven steps he is one step away from the starting point is

For a binomial distribution $B(n, p)$,if the mean $= 200$ and the standard deviation $= 10$,then the value of $n^2 + \frac{1}{p^2} + \frac{1}{q^2}$ is equal to: