Q.For the probability distribution of $X$ (the number of heads in two coin tosses) obtained in the previous example, find the mathematical expectation $E(X)$ and interpret its meaning.

The distribution is $P(X=0)=\tfrac{1}{4}$, $P(X=1)=\tfrac{1}{2}$, $P(X=2)=\tfrac{1}{4}$; $E(X)$ is found by multiplying each value by its probability and adding.

Using the distribution obtained earlier:

Cross-check: computing the same sum as fractions throughout, $0 \times \tfrac{1}{4} = 0$, $1 \times \tfrac{1}{2} = \tfrac{1}{2}$, $2 \times \tfrac{1}{4} = \tfrac{1}{2}$; adding, $0 + \tfrac{1}{2} + \tfrac{1}{2} = 1$. Both the decimal and fractional routes agree, confirming $E(X) = 1$.

Interpretation: $E(X) = 1$ means that if two coins were tossed a very large number of times and the number of heads recorded each time, the average number of heads per toss would settle very close to $1$ — which matches the intuitive symmetry of tossing two fair coins.

A common slip is to compute a simple average of the $x_i$ values (i.e., $(0+1+2)/3 = 1$) while ignoring that the probabilities are not all equal — this happens to give the same numerical answer here purely by coincidence, but the correct method always weights each value by its own probability, not by treating every value as equally likely.