Q.Compute $P(X=k)$ for the binomial distribution $B(n,p)$ where

A Bernoulli trial is a random experiment with exactly two mutually exclusive, exhaustive outcomes — success or failure — repeated independently with the success probability $p$ held fixed across repetitions.

Bernoulli distribution (Definition 11.10): with $X(\text{success})=1$, $X(\text{failure})=0$,

Mean $=p$, variance $=pq$ (derived in the Mean-&-Variance-of-Standard-Distributions concept).

Binomial distribution (Definitions 11.11-11.12): if $X$ counts the number of successes across $n$ independent Bernoulli trials, each with the same success probability $p$, then $X\sim B(n,p)$ with pmf

$\binom nx$ counts the number of ways to choose which $x$ of the $n$ trials were successes; each such arrangement has probability $p^xq^{n-x}$ by independence, and summing $\binom nx p^xq^{n-x}$ over $x=0,\dots,n$ recovers the binomial expansion $(p+q)^n=1^n=1$ — the reason for the name "binomial".

Working with a binomial pmf. A single probability is a direct substitution: $P(X=k)=\binom nk p^kq^{n-k}$. Compound events use the complement/sum rules already familiar from the pmf toolkit: $P(X\ge1)=1-P(X=0)$, $P(X\le1)=P(X=0)+P(X=1)$, "at least $j$ will NOT..." reframes as a second binomial variable $Y=n-X\sim B(n,q)$ counting the complementary outcome.

Recovering $n,p$ from a stated relation. Ratios such as $\dfrac{P(X=4)}{P(X=2)}$ or $9P(X=4)=P(X=2)$ collapse algebraically because $\binom nx$ and one power of $p,q$ cancel, typically leaving a clean relation like $q=cp$ for a constant $c$; combined with $p+q=1$ this pins down $p$ (and hence $q$) exactly, after which the full pmf, mean and variance follow.