Key Idea: A discrete random variable lists outcomes and their probabilities in a table. The IB tests two things on it: finding a missing probability (the column must add to 1) and the expected value — the long-run average. Almost always Paper 1, by hand.
🎲 The probability distribution
- the discrete random variable — it takes separate values
- the probability of each value; each is between 0 and 1, and they sum to 1
📐 The three things you'll be asked
| Asked for | What to do | Key step |
|---|---|---|
| An unknown probability or letter k | Add all the probabilities and set the total = 1, then solve. | Σ P(X = x) = 1, then substitute back for what's asked. |
| The expected value E(X) | Multiply each value by its probability and add them. | E(X) = Σ x·P(X = x) |
| Is a game fair? | Let X = net gain. Work out E(X). Fair when E(X) = 0. | Set E(X) = 0 to find a fair prize/cost. |
- each value the variable can take
- the mean — the long-run average value of X
✏️ IB-style worked examples
IB-style question — find k so the probabilities sum to 1
A discrete random variable X has P(X = x) = kx for x = 2, 4, 6, 8. Find the value of k, then state P(X = 6).
Step by step:
Add all four probabilities and set the total equal to 1.
Solve for k.
Substitute back to get the probability asked for.
k = 0.05; P(X = 6) = 0.3.
IB-style question — compute the expected value E(X)
The number of goals X a team scores in a match has this distribution: P(X = 0) = 0.2, P(X = 1) = 0.5, P(X = 2) = 0.2, P(X = 3) = 0.1. Find E(X), the expected number of goals.
Step by step:
Multiply each value by its probability.
Add the terms.
E(X) = 1.2 goals (a mean can be a value X never actually takes).
IB-style question — is the game fair?
At a stall you pay $3 to spin a wheel. You win $10 with probability 0.2, otherwise you win nothing. Find the expected net gain per play and state whether the game is fair.
Step by step:
Use the NET gain: +$7 if you win ($10 − $3 stake), −$3 if you lose.
Work it out.
Compare with 0. A fair game needs E(X) = 0.
Expected net gain = −$1.00 per play, so the game is not fair (it favours the stall).
Important: In a fair-game question X is the net gain, so subtract the cost to play. A $10 prize that cost $3 is a net win of $7, not $10 — and a loss is the stake itself as a negative. Then set E(X) = 0 for a fair game; a positive E(X) favours the player, a negative one favours the house.
Tap each card to reveal the answer.
What must every probability distribution satisfy? Each P between 0 and 1, and they sum to 1 (Σ P = 1).
P(X = x) = cx for x = 1, 2, 3, 4 — find c c(1+2+3+4) = 10c = 1, so c = 0.1.
X takes 1, 2, 3 with probabilities 0.5, 0.3, 0.2 — find E(X) 1(0.5)+2(0.3)+3(0.2) = 1.7.
When is a game fair? When the expected net gain is 0, i.e. E(X) = 0.
Can E(X) be a value X never takes? Yes — it's a long-run average, not an outcome (e.g. 1.2).
Exam tips
- Probabilities sum to 1 — set Σ P(X = x) = 1 to find an unknown, then substitute back.
- E(X) = Σ x·P(X = x): multiply each value by its probability, then add.
- E(X) need not be a value X can take — it's the long-run average.
- For a game, let X be the NET gain (subtract the stake); it is fair when E(X) = 0.
- This topic is non-calculator (Paper 1) — show the Σ working, not just the answer.