Q.Using the binomial distribution, find the mean and variance of $X$ for the following experiments.

Four named discrete distributions, in order of increasing generality, each with its own closed-form mean and variance:

• One-point distribution ($f(x_0)=1$): mean $=x_0$, variance $=0$ — no spread is possible when all mass sits at one value (matches Corollary 4 of the expectation properties, $V(b)=0$).
• Two-point distribution (values $x_1,x_2$ with probabilities $p,q=1-p$): mean $=px_1+qx_2$, variance $=pq(x_2-x_1)^2$; in the symmetric case $p=q=\tfrac12$ these become $\dfrac{x_1+x_2}{2}$ and $\dfrac{(x_2-x_1)^2}{4}$.
• Bernoulli distribution $X\sim\text{Ber}(p)$ (the two-point distribution specialised to $x_1=1,x_2=0$): mean $=p$, variance $=pq$; symmetric case ($p=\tfrac12$) gives mean $\tfrac12$, variance $\tfrac14$.
• Binomial distribution $X\sim B(n,p)$ (the sum of $n$ independent Bernoulli$(p)$ trials): mean $=np$, variance $=np(1-p)=npq$; symmetric case ($p=\tfrac12$) gives mean $\dfrac n2$, variance $\dfrac n4$.

Reading a problem "backwards" from mean/variance. A large family of exercises states $\mu=np$ and $\sigma^2=npq$ (or $\sigma$) numerically and asks for $n,p$ (and hence the full distribution) — divide the two given equations to eliminate $n$: $\dfrac{\sigma^2}{\mu}=\dfrac{npq}{np}=q$, giving $q$ (hence $p=1-q$) directly; then $n=\mu/p$. The same pair of equations, run the other way, instantly gives the standard deviation or variance once $n,p$ are known ($\sigma=\sqrt{np(1-p)}$).

Sanity checks. $0<p<1\Rightarrow0<q<1$, and $n$ must come out a positive integer — if solving the mean/variance pair yields a non-integer or negative $n$, recheck the algebra (a sign or ratio slip is the usual cause, since $q=\sigma^2/\mu$ must itself lie strictly between $0$ and $1$).