IB Maths AI SL Probability Distributions Paper 1 & 2 Expected value ~6 min read

Expected Values, E(X)

The expected value E(X) is the mean of a discrete random variable — the long-run average value if you repeated the experiment many, many times. You find it by multiplying each value by its probability and adding everything up. Its headline application: deciding whether a game is fair by comparing the expected prize to the cost of playing.

📘 What you need to know

E(X) is the mean of a random variable — the average outcome over many repeats, not necessarily a value X can actually take.
The formula: E(X) = ∑ x P(X = x) — multiply each value by its probability, then add. In the booklet.
Build the table first: you can only sum x P(X=x) once every probability is known — solve for unknowns first.
Symmetry shortcut: if the values AND probabilities are symmetric, E(X) is just the centre value — no calculation needed.
Fair game: a game is fair when the expected gain is exactly 0 — the expected prize equals the cost to play.
Expected gain = E(prize) − cost. Positive ⇒ player profits long-term; negative ⇒ player loses long-term.

Calculating E(X)

The expected value is a weighted average: each possible value of X is weighted by how likely it is. Think of it as the balance point of the distribution — the place a ruler with the probability “weights” would balance.

The expected value is the weighted average of the distribution — the point where it would balance if each probability were a physical weight. Here E(X) = 1.45, which sits between the values 1 and 2 and need not be an actual outcome of X.

Expected value of a discrete random variable E(X) = ∑ x P(X = x) multiply each value by its probability, then add them all

The symmetry shortcut

If a distribution is symmetric — the values are evenly spaced around a centre AND their probabilities mirror each other — then E(X) is simply that centre value. No multiplication needed.

Example: if X takes values 2, 5, 8, 11, 14 with probabilities 0.1, 0.2, 0.4, 0.2, 0.1, the values are symmetric about 8 and so are the probabilities. By symmetry, E(X) = 8 — the full calculation confirms it but isn’t necessary.

Fair games — the headline application

In a game, let the random variable be the player’s gain (positive for a win, negative for a loss). The expected gain is:

Expected gain in a game E(gain) = E(prize) − cost to play

Reading the result: if E(gain) > 0 the player profits in the long run; if E(gain) < 0 the player loses; if E(gain) = 0 the game is fair. To make a game fair, the cost to play must exactly equal the expected value of the prize.

🧭 Recipe — any expected value question

Build the complete table: all values of X and all probabilities. Solve for any unknowns using ∑P(X=x) = 1 first.
Check for symmetry: if values and probabilities mirror about a centre, E(X) is that centre — done.
Otherwise apply the formula: E(X) = ∑ x P(X = x) — multiply each value by its probability, then sum.
For a game, compute E(prize), then expected gain = E(prize) − cost.
Interpret: E(gain) = 0 fair · > 0 player profits · < 0 player loses. State it in words and in context.

Worked examples

WE 1

Basic expected value from a table

The number of pets X owned by a randomly chosen student has the distribution:

x: 0, 1, 2, 3 | P(X=x): 0.2, 0.3, 0.35, 0.15

Find E(X).

apply E(X) = ∑ x P(X = x) E(X) = 0(0.2) + 1(0.3) + 2(0.35) + 3(0.15) work through each term = 0 + 0.3 + 0.7 + 0.45 = 1.45 E(X) = 1.45 pets E(X) = 1.45 isn’t a value X can actually take — you can’t own 1.45 pets. It’s the long-run average across many students. That’s normal for an expected value.

WE 2

Find a missing probability, then E(X)

The distribution of X has one unknown probability p:

x: 1, 2, 3, 4 | P(X=x): 0.1, p, 0.4, 0.2

(a) Find p. (b) Find E(X).

(a) total rule first 0.1 + p + 0.4 + 0.2 = 1 0.7 + p = 1 ⇒ p = 0.3 (b) now E(X) = ∑ x P(X = x) = 1(0.1) + 2(0.3) + 3(0.4) + 4(0.2) = 0.1 + 0.6 + 1.2 + 0.8 = 2.7 (a) p = 0.3 · (b) E(X) = 2.7 you CANNOT compute E(X) until every probability is known. Always settle the missing probability with ∑P = 1 before touching the expected value formula.

WE 3

Use symmetry to find E(X)

The random variable X has the distribution:

x: 2, 5, 8, 11, 14 | P(X=x): 0.1, 0.2, 0.4, 0.2, 0.1

Find E(X), explaining your method.

check for symmetry values 2, 5, 8, 11, 14 — evenly spaced, centre 8 probs 0.1, 0.2, 0.4, 0.2, 0.1 — mirror about the centre distribution is symmetric ⇒ E(X) = centre value E(X) = 8 (check) 2(0.1)+5(0.2)+8(0.4)+11(0.2)+14(0.1) = 0.2+1+3.2+2.2+1.4 = 8 ✓ E(X) = 8 (by symmetry) when both the values and the probabilities mirror about a centre, the mean IS that centre. Spotting symmetry can save the whole calculation — but a quick check confirms it.

WE 4

E(X) from a probability function

The discrete random variable X has P(X = x) = x15 for x = 1, 2, 3, 4, 5. Find E(X).

build the table: P(X=x) = x/15 P-values: 1/15, 2/15, 3/15, 4/15, 5/15 apply E(X) = ∑ x P(X = x) = 1(1/15) + 2(2/15) + 3(3/15) + 4(4/15) + 5(5/15) = (1 + 4 + 9 + 16 + 25) / 15 = 55/15 = 11/3 E(X) = 11/3 ≈ 3.67 each term is x × (x/15) = x²/15, so E(X) becomes (sum of x²)/15. Spotting that pattern turns the whole sum into one tidy fraction.

WE 5

Is the game fair?

A charity stall charges $3 to play a game. The prize P (in dollars) has the distribution:

p: 0, 2, 5, 20 | P(P=p): 0.5, 0.3, 0.15, 0.05

(a) Find the expected prize. (b) Determine whether the game is fair.

(a) E(prize) = ∑ p P(P = p) = 0(0.5) + 2(0.3) + 5(0.15) + 20(0.05) = 0 + 0.6 + 0.75 + 1.0 = $2.35 (b) expected gain = E(prize) − cost = 2.35 − 3 = −0.65 E(gain) is negative (a) $2.35 · (b) NOT fair — expected LOSS of $0.65 per game a fair game needs E(gain) = 0. Here the player loses $0.65 on average per play — good for the charity, not for the player. A negative expected gain means the game favours the stall.

WE 6

Find the cost that makes a game fair

A game offers a prize P (in dollars) with the distribution:

p: 0, 4, 10, 50 | P(P=p): 0.55, 0.3, 0.1, 0.05

(a) Find the expected prize. (b) State the cost that would make the game fair. (c) If the organiser charges $4, find the player’s expected gain.

(a) E(prize) = ∑ p P(P = p) = 0(0.55) + 4(0.3) + 10(0.1) + 50(0.05) = 0 + 1.2 + 1.0 + 2.5 = $4.70 (b) fair ⇒ cost = E(prize) fair cost = $4.70 (c) charge is $4 E(gain) = E(prize) − cost = 4.70 − 4 = +0.70 (a) $4.70 · (b) $4.70 · (c) expected gain +$0.70 (player profits) at $4 the cost is BELOW the expected prize, so the player gains $0.70 per game on average — the game favours the player. To break even (fair), the organiser must charge exactly $4.70.

💡 Top tips

Settle unknowns first: use ∑P(X=x) = 1 to find any missing probability BEFORE computing E(X).
Check for symmetry: symmetric values + symmetric probabilities ⇒ E(X) is the centre. Big time-saver.
E(X) need not be a possible value — “1.45 pets” or “2.5 tails” are perfectly valid expected values.
For games: expected gain = E(prize) − cost. Fair ⇔ this equals 0.
Always interpret in context: state “expected loss of $0.65 per game”, not just “−0.65”.

⚠ Common mistakes

Computing E(X) with a missing probability: the formula needs the complete distribution — find unknowns first.
Adding the values without weighting: E(X) is ∑xP(X=x), NOT the plain average of the values.
Forgetting to subtract the cost: E(prize) alone isn’t the expected gain — you must subtract what the player paid.
Sign confusion on “fair”: fair means E(gain) = 0, not E(prize) = 0. Compare prize against cost.
Expecting E(X) to be a listed value: it’s an average — it can fall between the actual outcomes.

That completes the Probability Distributions chapter. You can now describe a discrete random variable fully — its distribution table, the total rule, cumulative probabilities — and find its mean with E(X) = ∑xP(X=x). The “fair game” idea is a favourite exam application, blending expected value with real-world decision-making. Next chapters build toward named distributions like the binomial and the normal, but the foundation — values, probabilities, and weighted averages — is exactly what you’ve built here.

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