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Contents · 0% complete
  1. Learning objectives
  2. Core idea
  3. Key formulae
  4. Worked examples
  5. Explore the graph
  6. Common mistakes
  7. Exam tip
  8. Guided practice
  9. Practice questions
  10. Challenge
  11. Summary
  12. Quick quiz
  13. Same topic, different board
  14. Past paper questions
Discrete random variables · Subtopic 1 of 3

Probability Distributions

A random variable turns the outcome of an experiment into a number. Listing each possible value with its probability gives a distribution table, and the probabilities must add up to 11.

Cambridge 9709 Paper 5Edexcel 9MA0 Paper 3Edexcel 9FM0 OptionAQA 7357 Paper 3AQA 7367 Paper 3≈ 30 minRevision 5 · updated 2026-10-27
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Section 1 of 14

Learning objectives

  • Understand a discrete random variable as a numerical outcome with a set of probabilities.
  • Draw up a probability distribution table from a situation.
  • Use ΣP(X=x)=1\Sigma P(X=x)=1 to find unknown probabilities or constants.
  • Find probabilities such as P(X≥2)P(X\ge2) from a table.
Pause and think

Two coins are tossed and XX is the number of heads. What values can XX take, and are they equally likely?

Section 2 of 14

Core idea

Start hereEvery outcome with its chance

Toss three coins and count heads: 00, 11, 22 or 33. Each count has a probability. Listing them all in a table gives a complete picture of the experiment: a probability distribution.

The big idea

A discrete random variable XX takes separate values xx, each with probability P(X=x)P(X=x). The table of values and probabilities is its distribution; the probabilities must sum to 11.

1. Building a distribution

Distribution of the total of two dice
246810120.050.10.15total xP(X = x)
From the 36-cell grid: P(X=2)=136P(X=2)=\tfrac1{36}, rising to P(X=7)=636P(X=7)=\tfrac6{36}, then falling. The bars add to 11.
Worked exampleHeads in three tosses

Find the distribution of XX = number of heads.

Goal
Table.
Strategy
List the 8 equally likely outcomes.
  1. HHH; HHT,HTH,THH; HTT,THT,TTH; TTTHHH;\ HHT,HTH,THH;\ HTT,THT,TTH;\ TTT
    Grouped by heads.
  2. P(0)=18, P(1)=38, P(2)=38, P(3)=18P(0)=\tfrac18,\ P(1)=\tfrac38,\ P(2)=\tfrac38,\ P(3)=\tfrac18
    Count.
AnswerAs above.
✓ CheckSum =1=1 ✓.
Worked exampleP(X=x)=kxP(X=x)=kx for x=1,2,3,4x=1,2,3,4

Find kk and P(X≥3)P(X\ge3).

Goal
kk; probability.
Strategy
Sum to 11.
  1. k(1+2+3+4)=1⇒k=110k(1+2+3+4)=1\Rightarrow k=\tfrac1{10}
    Total probability.
  2. P(X≥3)=3+410=0.7P(X\ge3)=\tfrac{3+4}{10}=0.7
    Add.
Answerk=0.1k=0.1; 0.70.7.
✓ Check0.1+0.2+0.3+0.4=10.1+0.2+0.3+0.4=1 ✓.
Common misconception<< vs ≤\le For discrete XX, P(X<3)=P(X≤2)P(X<3)=P(X\le2). Read inequalities carefully.
Pause and think

Two cards from 2 red and 3 black without replacement. Distribution of reds?

Summary

  • List all values with probabilities.
  • ∑P=1\sum P=1.
Key words
Random variable
A numerical outcome of a random process.
HistoryThe first probability textbook. In 1657 the Dutch scientist Christiaan Huygens published De ratiociniis in ludo aleae ("On reasoning in games of chance"), the first printed book on probability. It worked through dice and card problems by listing every possible payoff with its chance: probability distributions in all but name.
Fun factSpotting a loaded die. A fair die has the uniform distribution P(X=x)=16P(X=x)=\tfrac16. Casinos test dice by rolling them thousands of times and comparing the observed frequencies with this distribution; a die that shows six 20%20\% of the time instead of 16.7%16.7\% is quickly removed from play.
Section 3 of 14

Key formulae

Sum of probabilities∑all xP(X=x)=1\sum_{\text{all }x}P(X=x)=1
Probabilities of rangesP(X≥a)=∑x≥aP(X=x)=1−P(X<a)P(X\ge a)=\sum_{x\ge a}P(X=x)=1-P(X<a)
TrickList the values before the probabilities. First write the top row of the table: every value XX can take, smallest to largest. Then work out each probability, with a tree diagram if the experiment has stages. Finally check the bottom row adds to 11; if it doesn't, a value or a branch is missing.
Section 4 of 14

Worked examples

Example 1 · A formula0 / 2 steps
P(X=x)=k(x+1)P(X=x)=k(x+1) for x=0,1,2,3x=0,1,2,3. Find kk and P(X≥2)P(X\ge2).
    Example 2 · Without replacement0 / 3 steps
    A bag has 33 red and 22 blue counters. Two are taken without replacement and XX is the number of red counters. Find the distribution of XX.
      Example 3 · Two dice0 / 2 steps
      Two fair dice are thrown and DD is the (non-negative) difference between the scores. Find P(D=0)P(D=0) and P(D=1)P(D=1).
        Section 5 of 14

        Explore the graph

        A distribution given by a formula
        Loading the interactive graph…
        Try this
        1. The bars show P(X=x)=kxP(X=x)=kx for x=1,2,3,4x=1,2,3,4.
        2. For a valid distribution the bar heights must add to 11. Find the kk that does this.
        3. For k=0.2k=0.2 the heights add to 22. Why is that impossible?

        A formula becomes a distribution only once kk makes the probabilities sum to 11.

        In the real worldInsurance tables. Insurers model the number of claims a customer makes in a year as a random variable: perhaps P(X=0)=0.85P(X=0)=0.85, P(X=1)=0.12P(X=1)=0.12, P(X=2)=0.03P(X=2)=0.03. The whole premium calculation starts from a distribution table like this.
        Section 6 of 14

        Common mistakes

        IncorrectLeaving out a value, such as X=0X=0
        CorrectList every possible value
        The probabilities then cannot sum to 11.
        IncorrectP(X≥2)P(X\ge2) for values 00–33 =P(X=3)=P(X=3)
        CorrectP(X=2)+P(X=3)P(X=2)+P(X=3)
        "At least 22" includes 22.
        IncorrectAssuming the values are equally likely
        CorrectWork out each probability
        Two heads and one head are not equally likely.
        IncorrectA probability bigger than 11 or negative
        CorrectCheck the table
        Each probability must be in [0,1][0,1].

        Probabilities that add to 1.21.2

        Probabilities that add to exactly 11

        Section 7 of 14

        Exam tip

        Exam tipShow one probability in full. When asked to "show that P(X=1)=35P(X=1)=\tfrac35", write every factor of the calculation, including the "×2\times2" for the two orders. "Show that" questions give the answer, so the marks are entirely for the working.
        Section 8 of 14

        Guided practice

        SA-9709-P5-DV1-0012 marksL1 FoundationCalculatorDistributions
        The random variable XX has P(X=x)=k(x+1)P(X=x)=k(x+1) for x=0,1,2,3x=0,1,2,3. Find kk.

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        Salam Academy Original Practice · Proposed mark scheme
        Section 9 of 14

        Practice questions

        SA-9709-P5-DV-0012 marksL1 FoundationCalculatorDistributions
        The random variable XX has P(X=x)=kxP(X=x)=kx for x=1,2,3,4x=1,2,3,4. Find kk.

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        Salam Academy Original Practice · Proposed mark scheme
        SA-9709-P5-DV1-0021 markL1 FoundationCalculatorDistributions
        XX takes the values 11, 22 and 33 with probabilities 0.20.2, 0.50.5 and pp. Find pp.

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        Salam Academy Original Practice · Proposed mark scheme
        SA-9709-P5-DV-0021 markL1 FoundationCalculatorBuilding tables
        Three fair coins are thrown and XX is the number of heads. Find P(X=2)P(X=2).

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        Salam Academy Original Practice · Proposed mark scheme
        SA-9709-P5-DV1-0051 markL1 FoundationCalculatorDistributions
        XX takes the values 0,1,2,30,1,2,3 with probabilities 0.1, 0.3, 0.4, 0.20.1,\ 0.3,\ 0.4,\ 0.2. Find P(X≥2)P(X\ge2).

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        Salam Academy Original Practice · Proposed mark scheme
        SA-9709-P5-DV1-0031 markL1 FoundationCalculatorBuilding tables
        Two fair coins are tossed and XX is the number of tails. Find P(X=1)P(X=1).

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        Salam Academy Original Practice · Proposed mark scheme
        SA-9709-P5-DV1-0043 marksL2 StandardCalculatorBuilding tables
        A bag contains 33 red and 22 blue counters. Two are taken at random without replacement, and XX is the number of red counters taken. Find P(X=1)P(X=1) and P(X=2)P(X=2).

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        Salam Academy Original Practice · Proposed mark scheme
        SA-9709-P5-DV1-0062 marksL2 StandardCalculatorBuilding tables
        Two fair dice are thrown and DD is the difference between the scores (larger minus smaller, or 00 if equal). Find P(D=0)P(D=0) and P(D=1)P(D=1).

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        Salam Academy Original Practice · Proposed mark scheme
        SA-9709-P5-DV1-0072 marksL2 StandardCalculatorDistributions
        The random variable XX has P(X=x)=kxP(X=x)=\dfrac kx for x=1,2,3x=1,2,3. Find kk.

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        Salam Academy Original Practice · Proposed mark scheme

        More questions on every skill in this subtopic are in the question bank under Practise.

        Section 10 of 14

        Challenge

        SA-9709-P5-DV-0135 marksL4 ChallengingCalculatorBuilding tables
        Two fair dice are thrown and XX is the larger of the two scores (or the common score if they are equal). Find P(X=4)P(X=4) and E(X)E(X).

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        Salam Academy Original Practice · Proposed mark scheme
        Section 11 of 14

        Summary

        • A discrete random variable takes separate values, each with a probability.
        • List every value, then every probability; they sum to 11.
        • Use the sum to find unknown constants.
        • Read ranges like P(X≥2)P(X\ge2) by adding the right columns.
        Section 12 of 14

        Quick quiz

        Q1 · SA-9709-P5-DV-Q011 markL1 FoundationCalculatorDistributions
        In any probability distribution, the probabilities
        Choose one option

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        Salam Academy Original Practice · Proposed mark scheme
        Q2 · SA-9709-P5-DV1-0081 markL1 FoundationCalculatorDistributions
        Which of these could be a probability distribution for XX taking the values 0,1,20,1,2?
        Choose one option

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        Salam Academy Original Practice · Proposed mark scheme
        Section 13 of 14

        Same topic, different board

        Specification
        7357 A-level Mathematics
        Where
        Section O, Statistical distributions
        Expected
        Discrete random variables given in a table or by a simple formula.
        Calculator
        Permitted on all papers; a calculator with statistical distribution functions is expected.
        Typical style
        Questions may draw on the pre-released large data set.
        Open in
        7357 · Paper 37367 · Paper 3
        Section 14 of 14

        Past paper questions

        8 questions (48 marks) from real exams test this subtopic. Try each one under timed conditions, then mark it with the official mark scheme.

        1. 9709/52 · Oct/Nov 2025Question 27 marks

          Probability distribution from selections without replacement, and its variance.

        2. 9709/52 · May/June 2025Question 13 marks

          Probability distribution for the number of tails from three biased coins.

        3. 9709/52 · Feb/March 2025Question 16 marks

          Distribution of heads from three biased coins, then the variance.

        4. 9709/52 · Oct/Nov 2024Question 36 marks

          Probability distribution of a coin-and-dice variable and its variance.

        5. 9709/52 · May/June 2024Question 57 marks

          Probability distribution, expectation and variance of the total of two coins.

        6. 9709/52 · May/June 2023Question 16 marks

          Probability distribution from a formula with an unknown constant, then mean and variance.

        Show 2 more
        1. 9709/52 · Oct/Nov 2022Question 57 marks

          Fair and biased coins: a conditional probability and a probability distribution.

        2. 9709/52 · May/June 2022Question 26 marks

          Distribution of the sum of two rolls of a non-standard die, mean and variance.

        All 9709 past paper questions by topic

        Salam Academy original content · Content status: reviewed and published. Syllabus mapping: AQA 7367, First teaching 2017 (official specification).