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 1

Verify 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 2

Find 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 3

Distribution 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 4

Discrete 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 5

Solve for an unknown using algebra

The discrete random variable Z has distribution:

z	1	2	3	4
P(Z = z)	2a	3a	a	0.1

(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 6

Real-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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