A table of outcomes and their probabilities: A discrete random variable X takes separate values, each with a probability.
A probability distribution lists them in a table.
The probabilities must be valid (each between 0 and 1) and sum to 1.
A discrete probability distribution: the bar heights are the probabilities (they sum to 1); E(X) is the balance point.
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
IB-style question — is it valid?
A variable X has P(X = 1) = 0.1, P(X = 2) = 0.3, P(X = 3) = 0.4, P(X = 4) = 0.2.
Show that this is a valid probability distribution.
Step by step
- Each probability is between 0 and 1, and add them.
- Since they total 1, it is valid.
Final answer
All probabilities are valid and sum to 1, so it is a valid distribution.
The total is always 1: If the probabilities don't add to 1, either a value is missing or there's an error.
IB-style question — mode and a conditional
A discrete random variable X has P(X = 0) = 0.1, P(X = 1) = 0.2, P(X = 2) = 0.5, P(X = 3) = 0.2.
(a) Write down the mode.
(b) Find P(X ≥ 2 | X ≥ 1).
Step by step
- (a) The mode is the value with the highest probability.
- (b) Conditional probability — restrict to X ≥ 1 and take the X ≥ 2 part of it.
Final answer
(a) mode = 2. (b) 7⁄9 ≈ 0.778.
The mode is the tallest bar (X = 2). For the conditional, the denominator is the total probability of X ≥ 1 and the numerator the X ≥ 2 part.
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
Free preview
This is the free notes preview
You're reading the free notes. Aimnova Pro unlocks the full study experience — and you can try it with your first topic free to keep:
- FlashcardsLock in vocabulary and key terms with spaced repetition.
- Practice questionsAnswer exam-style questions and get instant AI marking.
- Mock exams & past-paper vaultSit full mocks and see exactly how examiners award marks.
- Personalised study planA daily plan built around your exam date and weak areas.
Use Σ P = 1 to solve for the unknown: If a probability (or a parameter inside one) is unknown, set the sum of all probabilities equal to 1 and solve.
IB-style question — find k
A distribution has P(X = x) = kx for x = 1, 2, 3, 4.
Find k, then P(X = 3).
Step by step
- Sum the probabilities and set equal to 1.
- Solve, then substitute.
Final answer
k = 0.1; P(X = 3) = 0.3.
Substitute back: After finding the parameter, substitute to get the actual probability the question asks for.
See how examiners mark answers
Access past paper questions with model answers. Learn exactly what earns marks and what doesn't.
Sum of value × probability: The expected value (mean) of X is E(X) = Σ x·P(X = x) — multiply each value by its probability and add.
It's the long-run average value of X.
IB-style question — find E(X)
X has P(X = 1) = 0.1, P(X = 2) = 0.2, P(X = 3) = 0.3, P(X = 4) = 0.4.
Find E(X).
Step by step
- Multiply each value by its probability.
- Add.
Final answer
E(X) = 3.
E(X) need not be a possible value: The mean can be a value X never takes (e.g. 2.7) — it's an average, not an outcome.
E(X) = the long-run average gain: For a game, let X be the net gain.
E(X) is the average gain per play.
A game is fair when E(X) = 0 (no expected gain or loss).
Set E(X) = 0 to find a fair prize.
IB-style question — make it fair
You roll a fair die.
You win $w if it shows a six, otherwise you lose $2.
Find the prize w that makes the game fair.
Step by step
- Net gain: +w with prob 1/6, −2 with prob 5/6. Set E(X) = 0.
- Solve for w.
Final answer
A prize of $10 makes the game fair.
Use the NET gain: Include the cost or loss in X.
If a $5 prize costs $5 to enter, the net 'win' is $0, not $5.