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.
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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.
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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.
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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.