Q.Two coins are to be flipped. The first coin will land on heads with probability $0.6$, the second with probability $0.5$. Assume that the results of the flips are independent, and let $X$ equal the total number of heads that result. The value of $E(X)$ is

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. …