A probability distribution can be shown as a table: the top row lists the values of X, the bottom row lists their probabilities. The bottom row must sum to 1. If a probability is unknown, call it a letter and solve.
P(X ≤ k) — “at most
k“: add probabilities for all values up to and INCLUDING
k.
P(X > k) = 1 − P(X ≤ k) — the complement is often the fast route.
P(X ≥ k) = 1 − P(X < k).
Watch the boundary carefully: ≤ includes k, < excludes it. “At most 3” includes 3; “fewer than 3” stops at 2.
🧭 Recipe — any discrete distribution question
- Build the table: list every value of X and its probability. If given a function, substitute each value to get the probabilities.
- Apply the total rule: set ∑ P(X = x) = 1. If there’s an unknown, this is your equation — solve it.
- Translate the question’s words into an inequality (≤, <, ≥, >) — mind whether the boundary value is included.
- Add the relevant probabilities. For “more than” or “at least”, consider 1 − (the complement) as a shortcut.
- Sanity check: every probability is between 0 and 1, and any answer for a range can’t exceed 1.
Worked examples
WE 1Find a missing probability using the total rule
The discrete random variable X has the distribution below, where p is unknown.
x: 1, 2, 3, 4 | P(X=x): 0.15, 0.3, p, 0.25
Find the value of p.
all probabilities must add to 1
0.15 + 0.3 + p + 0.25 = 1
collect the known values
0.7 + p = 1
p = 1 − 0.7 = 0.3
p = 0.3
the total rule ∑P(X=x) = 1 is the single most useful tool here — any one missing probability can always be found by subtracting the rest from 1.
WE 2Solve for an unknown constant in a table
The distribution of X is given in terms of a constant k:
x: 1, 2, 3, 4 | P(X=x): k, 2k, 3k, 4k
(a) Find k. (b) State P(X = 3).
(a) apply the total rule
k + 2k + 3k + 4k = 1
10k = 1
k = 1/10
(b) P(X = 3) = 3k
P(X = 3) = 3 × 1/10 = 3/10
(a) k = 1/10 · (b) P(X = 3) = 3/10
collect like terms first: k + 2k + 3k + 4k = 10k. Then one division gives k. Always check: 1/10 + 2/10 + 3/10 + 4/10 = 10/10 = 1 ✓.
WE 3Distribution given as a function
The discrete random variable X has probability distribution function P(X = x) = cx for x = 1, 2, 3, 4, 5, and 0 otherwise. (a) Find c. (b) Find P(X ≥ 4).
(a) substitute each x to get probabilities
P-values: c, 2c, 3c, 4c, 5c
total rule: c + 2c + 3c + 4c + 5c = 1
15c = 1 ⇒ c = 1/15
(b) P(X ≥ 4) = P(X=4) + P(X=5)
= 4c + 5c = 9c
= 9 × 1/15 = 9/15 = 3/5
(a) c = 1/15 · (b) P(X ≥ 4) = 3/5
“function” form just means: substitute the valid x-values to build the table, then proceed as normal. P(X ≥ 4) includes 4 itself (the ≥ sign).
WE 4Cumulative probabilities — reading the inequality
The number of goals X scored by a team in a match has the distribution:
x: 0, 1, 2, 3, 4 | P(X=x): 0.1, 0.25, 0.3, 0.2, 0.15
Find (a) P(X ≤ 2), (b) P(X > 2), (c) P(1 ≤ X ≤ 3).
(a) P(X ≤ 2) = P(0) + P(1) + P(2)
= 0.1 + 0.25 + 0.3 = 0.65
(b) P(X > 2) = 1 − P(X ≤ 2)
= 1 − 0.65 = 0.35
(c) P(1 ≤ X ≤ 3) = P(1) + P(2) + P(3)
= 0.25 + 0.3 + 0.2 = 0.75
(a) 0.65 · (b) 0.35 · (c) 0.75
(b) uses the complement — “more than 2” is the opposite of “at most 2”, so subtract from 1. For (c), the ≤ signs mean BOTH endpoints (1 and 3) are included.
WE 5A quadratic probability function
The discrete random variable X has P(X = x) = kx2 for x = 1, 2, 3, 4, and 0 otherwise. (a) Show that k = 130. (b) Find P(X ≤ 2).
(a) substitute x = 1, 2, 3, 4 into kx²
P-values: k(1), k(4), k(9), k(16)
= k, 4k, 9k, 16k
total rule: k + 4k + 9k + 16k = 1
30k = 1 ⇒ k = 1/30 ✓
(b) P(X ≤ 2) = P(X=1) + P(X=2)
= k + 4k = 5k
= 5 × 1/30 = 5/30 = 1/6
(a) k = 1/30 · (b) P(X ≤ 2) = 1/6
for kx², square the x-value FIRST, then multiply by k. The squares 1, 4, 9, 16 sum to 30 — that’s where the 30 comes from.
WE 6Discrete uniform distribution
A fair eight-sided die has faces numbered 1 to 8. Let X be the number rolled. (a) State the distribution of X. (b) Find P(X is a multiple of 3). (c) Find P(X > 5).
(a) all 8 faces equally likely
discrete uniform: P(X = x) = 1/8 for x = 1,…,8
(b) multiples of 3 in 1 to 8: {3, 6}
2 favourable values, each prob 1/8
P(multiple of 3) = 2/8 = 1/4
(c) X > 5 means {6, 7, 8}
3 values, each prob 1/8
P(X > 5) = 3/8
(a) uniform, P = 1/8 · (b) 1/4 · (c) 3/8
a discrete uniform distribution makes probabilities easy: just count the favourable values and divide by n. Here n = 8. Note “> 5” excludes 5 itself.
💡 Top tips
- Always build the table first — even for function-form questions. A clear table prevents counting slips.
- ∑P(X=x) = 1 is your equation for ANY unknown — one missing probability or one constant.
- Underline the inequality words: “at most” ≤ · “fewer than” < · “at least” ≥ · “more than” >.
- Use the complement for “more than” / “at least” when there are many values above k: 1 − P(the rest).
- Discrete uniform: spot “fair” / “equally likely” — each of n values has probability 1n.
⚠ Common mistakes
- Probabilities not summing to 1: if your table doesn’t total 1, there’s an error — recheck before going further.
- Boundary errors: P(X ≤ 3) includes 3; P(X < 3) does not. Mixing these up loses easy marks.
- Forgetting to square / apply the function: for P(X=x) = kx2, use x2 not x.
- Including invalid values: P(X = k) = 0 if k is not one of the listed values — don’t invent a probability for it.
- Treating a DRV as continuous: discrete variables take only the listed separate values — there is no P(X = 2.5) if 2.5 isn’t in the table.
Next up: Expected Values, E(X) — the “mean” of a random variable. You’ll multiply each value by its probability and add: E(X) = ∑xP(X=x). This powers the classic exam question of deciding whether a game is fair — comparing the expected prize against the cost to play.
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