IB Maths AI HLProbability DistributionsPaper 1 & 2~7 min read
Discrete Probability Distributions
A discrete random variable (DRV) is a quantity whose value depends on a random event and can only take certain separate values — usually a count, like the number of heads in 20 flips. A probability distribution lists every value the variable can take alongside its probability. The one rule that drives nearly every question: all the probabilities add up to 1. Get that, build a table, and you can answer anything the IB throws at you.
📘 What you need to know
A discrete random variable X takes separate (often counting) values. P(X = x) = the probability X equals x.
Capitals for the variable (X), lower case for outcomes (x).
A probability distribution lists all values with their probabilities — as a table or a function (pdf).
All probabilities sum to 1: ∑P(X = x) = 1. This is your main equation.
Unknown probabilities: use algebra, then solve with ∑P = 1.
Discrete uniform: n values each with probability 1n.
If k isn’t a possible value, P(X = k) = 0.
Inequalities: add up the probabilities of all values satisfying the condition (≤, <, ≥, >).
The golden rule: probabilities sum to 1
Every distribution must have probabilities that total exactly 1. This is how you find unknown values and check your work.
The defining property
∑ P(X = x) = 1
add every probability — they must total 1
a distribution as a vertical line graph
Each value gets a line whose height is its probability. Here ¼ + ⅛ + ⅛ + ½ = 1 ✓
Discrete uniform: if all n values are equally likely, each has probability 1n — e.g. a fair dice: each face has probability 16.
Building a table from a function
If the distribution is given as a function, substitute each value to get its probability, then lay it out in a table.
🧭 Recipe — function → table → unknowns
Substitute each value of x into the function to get each probability.
Put them in a table (values across the top, probabilities below).
Sum to 1: add all the probabilities and set the total equal to 1.
Solve for any unknown constant (e.g. k).
Inequality probabilities
For P(X ≤ k) and friends, just add the probabilities of all the values that satisfy the condition. Watch whether k itself is included.
Wording
Symbol
Includes k?
at most / no greater than
X ≤ k
yes
fewer than / less than
X < k
no
at least / no fewer than
X ≥ k
yes
more than / greater than
X > k
no
🧠 Memory aid — “or equal? then include it”
The line under the inequality (≤, ≥) means “or equal to“, so k is included. No line (<, >) means k is excluded. And for the opposite tail, use the complement: P(X > k) = 1 − P(X ≤ k).
Worked examples
WE 1
Find an unknown constant
P(X = x) = kx² for x = −3, −1, 2, 4 (and 0 otherwise). Show that k = 130.
substitute each x into kx²x=−3: 9k, x=−1: k, x=2: 4k, x=4: 16kprobabilities sum to 19k + k + 4k + 16k = 130k = 1k = 1/30 ✓always build the table first, then use ∑P = 1.
WE 2
Calculate an inequality probability
Using k = 130 from WE 1, calculate P(X ≤ 3).
find each probability (kx²)x=−3: 9/30 = 3/10x=−1: 1/30x=2: 4/30 = 2/15X ≤ 3 means x = −3, −1, 2 (not 4)P(X ≤ 3) = 3/10 + 1/30 + 2/15P(X ≤ 3) = 7/15only add the values that satisfy the condition.
WE 3
Unknown in a table
A DRV has P(X = x): 0.1, 0.3, a, 0.2 for x = 1, 2, 3, 4. Find a.
all probabilities sum to 10.1 + 0.3 + a + 0.2 = 10.6 + a = 1a = 0.4the missing probability fills the gap to 1.
WE 4
Use the complement
For the table in WE 3 (0.1, 0.3, 0.4, 0.2 for x = 1, 2, 3, 4), find P(X > 1).
is 5 a possible value?no — X only takes 1, 2, 3, 4P(X = 5) = 0any value outside the distribution has probability 0.
💡 Top tips
Always build a table first — values on top, probabilities below.
∑P(X = x) = 1 finds unknowns and checks your work.
≤ and ≥ include k; < and > exclude it. Read the wording.
Use the complement (1 − P) for “more than” or “at least one” style questions.
P(X = k) = 0 if k isn’t a value the variable can take.
Discrete uniform → 1n for each of n equally likely values.
⚠ Common mistakes
Probabilities not summing to 1. Always check — it’s the defining property.
Including or excluding k wrongly in an inequality. The line means “or equal”.
Adding the wrong values for ≤ / < — list which x satisfy the condition first.
Giving a non-zero probability to a value the variable can’t take.
Forgetting to substitute each value into a pdf before summing.
Negative or >1 probabilities — every P must be between 0 and 1.
Next up — Expected Values E(X). A distribution tells you all the probabilities; the expected value squeezes them into a single “average” outcome. You’ll compute E(X) = ∑xP(X = x), use symmetry as a shortcut, and decide whether a game is fair.
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