Expectation of a discrete random variable

Probability distributions

A table lists each value x and P(X = x). The probabilities must add up to 1, which is how you find a missing one.

Expectation and variance

• E(X) = sum of x P(X = x)
• E(X²) = sum of x² P(X = x)
• Var(X) = E(X²) - [E(X)]²

Linear changes

E(aX + b) = aE(X) + b and Var(aX + b) = a² Var(X). Adding b shifts the mean but does not change the spread.

Worked example

X takes 1, 2, 3 with probabilities 0.2, 0.5, 0.3. Find E(X).

1. E(X) = 1(0.2) + 2(0.5) + 3(0.3)
2. = 0.2 + 1.0 + 0.9

Answer: E(X) = 2.1

Check yourself

1. X takes 0, 1, 2 with P = 0.3, 0.45, p. Find p.
   Show answer

   0.25. Probabilities add to 1: p = 1 - 0.75 = 0.25.

2. X takes 2 and 5 with probabilities 0.4 and 0.6. Find E(X).
   Show answer

   3.8. E(X) = 0.8 + 3 = 3.8.

3. E(X) = 3. Find E(2X + 1).
   Show answer

   7. E(2X + 1) = 2(3) + 1 = 7.

4. Var(X) = 4. Find Var(3X - 2).
   Show answer

   36. Var(3X - 2) = 3² x 4 = 36.

5. E(X) = 2 and E(X²) = 5. Find Var(X).
   Show answer

   1. Var(X) = 5 - 4 = 1.

Practise with instant feedback

Sources

• Pearson Information Manual 2026/2027: International A level

Last updated 7 November 2026. Spotted a mistake? Email alevel@lifereadysociety.com.