Q.Two balls are drawn in succession without replacement from an urn containing four red balls and three black balls. Let $X$ be the number of red balls drawn. Find the probability mass function and the mean for $X$.

Mean (Definition 11.8): for a random variable $X$ with pmf/pdf $f(x)$,

$E(X)$ generalises the plain numerical average, weighting each value by its true probability rather than by $\tfrac1n$; it need not be a value $X$ can actually take, and is best read as the long-run average over many repetitions. Theorem 11.3 extends this to any function $g(X)$: $E(g(X))=\sum_x g(x)f(x)$ or $\int g(x)f(x)\,dx$; taking $g(X)=X^k$ gives the $k$-th moment $E(X^k)$.

Variance (Definition 11.9): $V(X)=E\big((X-E(X))^2\big)$, with the far more usable computing form

Standard deviation is $\sigma=\sqrt{V(X)}$; both are always $\ge0$. A smaller $\sigma^2$ means values cluster tightly around the mean; a larger $\sigma^2$ means they scatter more widely — even distributions sharing the same mean can differ sharply here.

Three linearity laws (for constants $a,b$): $E(aX+b)=aE(X)+b$ (so $E(aX)=aE(X)$ and $E(b)=b$); $V(X)=E(X^2)-(E(X))^2$ (restated); and $V(aX+b)=a^2V(X)$ (so $V(aX)=a^2V(X)$ and $V(b)=0$). These make quick work of a shifted/scaled random variable — e.g. a net "winning amount" that is a linear function of a raw count — without recomputing the distribution from scratch.

Worked technique. For a discrete $X$: tabulate $x$, $f(x)$, $xf(x)$, $x^2f(x)$; sum the last two columns to get $E(X)$ and $E(X^2)$ directly, then apply $V(X)=E(X^2)-(E(X))^2$. For a continuous $X$: compute $E(X)=\int xf(x)\,dx$ and $E(X^2)=\int x^2f(x)\,dx$ over the support, then the same variance formula.