Two cards are drawn successively with replacement from a well-shuffled pack of $52$ cards. Let $X$ denote the random variable of the number of kings obtained in the two drawn cards. Then $P(X=1) + P(X=2)$ equals:

If $X \sim B(4, p)$ and $P(X=0)=\frac{16}{81}$,then $P(X=4)=$

If $8$ coins are tossed simultaneously,what is the probability of getting at least $6$ heads?

Let $C_1$ and $C_2$ be two biased coins such that the probabilities of getting a head in a single toss are $\frac{2}{3}$ and $\frac{1}{3}$,respectively. Suppose $\alpha$ is the number of heads that appear when $C_1$ is tossed twice,independently,and $\beta$ is the number of heads that appear when $C_2$ is tossed twice,independently. Then,the probability that the roots of the quadratic polynomial $x^2 - \alpha x + \beta$ are real and equal is:

If $15$ coins are tossed,then the probability of getting $10$ heads is:

Suppose $X$ follows a binomial distribution with parameters $n$ and $p$,where $0 < p < 1$. If $\frac{P(X=r)}{P(X=n-r)}$ is independent of $n$ for every $r$,then $p$ is equal to