SL 4.7—Discrete random variables

All probabilities in a valid distribution must sum to 1: $\sum P(X=x)=1$.

$0.15+0.25+m+0.30=1$.

$0.70+m=1$.

$m=1-0.70=0.30$.

$0.25$ is simply one of the given probabilities copied by mistake, not the result of solving the equation.

$0.10$ would result from adding an extra 0.20 by error, e.g. mis-summing the given values.

$0.20$ could come from forgetting to add one of the terms (e.g. omitting 0.25) before subtracting from 1.

Only $m=0.30$ makes the total probability equal 1.

$E(X)=\sum x_i{p}_i$.

$E(X)=0(0.1)+1(0.4)+2(0.4)+3(0.1)$.

$=0+0.4+0.8+0.3$.

$E(X)=1.5$.

$1.0$ would result from omitting the $2(0.4)=0.8$ term.

$2.0$ overestimates by treating all outcomes as equally weighted average of the x-values only (simple mean of 0,1,2,3 rounded incorrectly).

$1.25$ could come from a division error such as dividing the correct sum by an incorrect number of terms.

Careful term-by-term summation gives $E(X)=1.5$.