IB Maths AA HL
Topic 4 — Statistics & Probability
Paper 1 & 2
~7 min read
Discrete Probability Distributions
A discrete random variable takes a countable list of values, each with a fixed probability. A probability distribution shows those values alongside their probabilities — and the probabilities must sum to 1. Once you can build that table, every “at least”, “at most”, “fewer than” question collapses to adding the right rows.
📘 What you need to know
- Discrete random variable (DRV): a variable that takes a countable set of values — usually a finite list.
- Notation: P(X = x) is the probability that X equals x; capitals for variables, lowercase for outcomes.
- Validity rule: ∑ P(X = x) = 1. Probabilities must sum to exactly 1.
- Two formats: a table of values vs probabilities, OR a function P(X = x) = f(x).
- If k isn’t a possible value: P(X = k) = 0.
- Cumulative: P(X ≤ k) = sum of probabilities for every value at most k.
- Discrete uniform: n equally likely values, each with probability 1/n.
- Inequality language: “at least k” → ≥; “at most k” → ≤; “fewer than k” → <; “more than k” → >.
The probability distribution table
The cleanest way to capture a discrete distribution is a two-row table: values on top, probabilities below. From it you can read off any P(X = x) directly and add probabilities to handle inequalities.
Validity check — every distribution must satisfy this
∑ P(X = x) = 1
If the distribution is given by a function P(X = x) = f(x), substitute each allowed x to populate the table. If a probability is unknown (often labelled k or a), use the validity rule to solve for it.
Reading probabilities from the table
P(X = k)
read it off
single cell — or 0 if k isn’t in the list
P(X ≤ k)
add up to and including k
includes the boundary k
P(X < k)
add strictly below k
excludes the boundary
P(X ≥ k)
1 − P(X < k)
faster than summing many cells
Boundary trap: “fewer than 4” means X < 4, so the value 4 is NOT included. “At most 4” means X ≤ 4 and DOES include 4. Read the wording carefully — a single misread loses easy marks.
🧭 Recipe — solve any distribution question
- List every value the random variable can take.
- Build the table: values on top, probabilities below (substitute into the function if given).
- Apply ∑P = 1 as a validity check, or to solve for any unknown.
- Mark the cells that satisfy the inequality you’re asked about.
- Add the marked probabilities — and check the answer is between 0 and 1.
Worked examples
WE 1Verify a distribution and find a cumulative probability
The discrete random variable Y has the following distribution:
| y | 1 | 2 | 3 | 4 | 5 |
|---|
| P(Y = y) | 0.15 | 0.20 | 0.30 | 0.25 | 0.10 |
|---|
(a) Verify that this is a valid probability distribution. (b) Find P(Y ≥ 3).
(a) Apply ∑P = 1
0.15 + 0.20 + 0.30 + 0.25 + 0.10 = 1.00 ✓
(b) Add cells where y ≥ 3
P(Y ≥ 3) = 0.30 + 0.25 + 0.10 = 0.65
(a) Valid; (b) P(Y ≥ 3) = 0.65
always do the sum-check first — it’s worth its own mark
WE 2Find an unknown probability using ∑P = 1
The discrete random variable W has distribution:
| w | 0 | 1 | 2 | 3 |
|---|
| P(W = w) | 0.2 | k | 0.35 | 0.15 |
|---|
(a) Find k. (b) Hence find P(W < 2).
(a) Apply ∑P = 1
0.2 + k + 0.35 + 0.15 = 1
k = 1 − 0.7 = 0.3
(b) “Fewer than 2” means w = 0 or 1 (NOT including 2)
P(W < 2) = P(W = 0) + P(W = 1) = 0.2 + 0.3 = 0.5
(a) k = 0.3; (b) P(W < 2) = 0.5
“fewer than k” is the strict inequality — the value k itself is excluded
WE 3Distribution given as a function
The discrete random variable X has probability distribution given by
P(X = x) = c(x + 1) for x = 0, 1, 2, 3, 4 (0 otherwise)
(a) Find the value of c. (b) Find P(X > 2).
(a) Substitute each x into c(x+1) and build a table
P(0) = c, P(1) = 2c, P(2) = 3c, P(3) = 4c, P(4) = 5c
∑P = c(1+2+3+4+5) = 15c = 1
c = 1/15
(b) “Greater than 2” means x = 3 or 4
P(X > 2) = 4c + 5c = 9c = 9/15 = 3/5
(a) c = 1/15; (b) P(X > 2) = 3/5
when given a function, ALWAYS write out the table first — it prevents algebra slips
WE 4Discrete uniform distribution
A fair eight-sided die labelled 1 to 8 is rolled once. Let X be the number shown. (a) State the probability distribution. (b) Find P(X is even). (c) Find P(2 ≤ X ≤ 6).
(a) Discrete uniform: each of 8 values equally likely
P(X = x) = 1/8 for x = 1, 2, …, 8
(b) Even values: {2, 4, 6, 8} — 4 outcomes
P(X even) = 4/8 = 1/2
(c) Values from 2 to 6 inclusive: {2, 3, 4, 5, 6} — 5 outcomes
P(2 ≤ X ≤ 6) = 5/8
(b) 1/2; (c) 5/8
discrete uniform shortcut: just count favourable values ÷ total values
WE 5Solve for an unknown using algebra
The discrete random variable Z has distribution:
(a) Find the value of a. (b) Find P(Z ≤ 2).
(a) Apply ∑P = 1
2a + 3a + a + 0.1 = 1
6a = 0.9
a = 0.15
(b) Add cells where z ≤ 2
P(Z ≤ 2) = 2a + 3a = 5a
= 5(0.15) = 0.75
(a) a = 0.15; (b) P(Z ≤ 2) = 0.75
leave probabilities in terms of a until the final substitution — cleaner working
WE 6Real-world distribution — at least, at most, impossible value
The number of pets N per household in a survey is given by:
| n | 0 | 1 | 2 | 3 | 4 |
|---|
| P(N = n) | 0.30 | 0.40 | 0.20 | 0.07 | 0.03 |
|---|
Find: (a) P(at least 2 pets); (b) P(at most 3 pets) using the complement; (c) P(N = 5).
(a) “At least 2” means n = 2, 3, or 4
P(N ≥ 2) = 0.20 + 0.07 + 0.03 = 0.30
(b) “At most 3” means NOT (n = 4) — use complement
P(N ≤ 3) = 1 − P(N = 4) = 1 − 0.03 = 0.97
(c) 5 is NOT a value the variable can take
P(N = 5) = 0
(a) 0.30; (b) 0.97; (c) 0
complement saves time when the “missing” side has fewer cells
💡 Top tips
- Always sum-check first: ∑P = 1 catches errors and finds unknowns in one step.
- Translate words to inequalities early: “at least 3” → ≥ 3, “fewer than 4” → < 4, “more than 5” → > 5.
- Write out the table even when given a function — it eliminates substitution errors.
- Use the complement when the “other side” has fewer cells: P(X ≥ k) = 1 − P(X < k).
- Discrete uniform: don’t compute — just use favourable count ÷ total count.
⚠ Common mistakes
- Forgetting to check ∑P = 1 — leaves an invalid distribution unflagged.
- Including or excluding the boundary wrongly — “at most 5” includes 5; “fewer than 5” doesn’t.
- Treating P(X = k) for an impossible k as nonzero — it’s exactly 0.
- Missing values when populating the table from a function — list every x in the domain before computing.
- Mixing up “fewer than 5” with “at most 5” — different inequalities, different answers.
Next: Mean & Variance of a discrete random variable. Once you have the distribution table, computing E(X) and Var(X) is just two short summations away — and they’re foundational for the rest of the topic.
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