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

Suppose $X$ has a binomial distribution $B(6, 1/2)$. Show that $X=3$ is the most likely outcome.
(Hint: $P(X=3)$ is the maximum among all $P(x_i)$,where $x_i = 0, 1, 2, 3, 4, 5, 6$)

In a workshop,there are five machines and the probability of any one of them to be out of service on a day is $\frac{1}{4}$. If the probability that at most two machines will be out of service on the same day is $\left(\frac{3}{4}\right)^{3} k$,then $k$ is equal to

$A$ bag contains $30$ white balls and $10$ red balls. $16$ balls are drawn one by one randomly from the bag with replacement. If $X$ is the number of white balls drawn,then $\left( \frac{\text{mean of } X}{\text{standard deviation of } X} \right)$ is equal to

The random variable $X$ follows a binomial distribution $B(n, p)$ for which the difference between the mean and the variance is $1$. If $2 P(X=2) = 3 P(X=1)$,then $n^2 P(X > 1)$ is equal to $......$.

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 at least one ball will bear the $'Y'$ mark.