Discrete Probability Distributions

Random variables, probability distributions, and probability mass functions

SPEC 4.1 • STATISTICAL DISTRIBUTIONS

Random Variables

Definition

A random variable is a variable whose value depends on the outcome of a random event. The value is not known until the experiment is carried out.

Notation

Random variables: capital letters (X, Y, Z)

Specific values: lowercase letters (x, y, z)

P (X = x)

means "probability X takes value x"

Sample Space

The sample space is the set of all possible values the random variable can take.

Example: For a die, sample space = {1, 2, 3, 4, 5, 6}

Discrete vs Continuous

A variable is discrete if it can only take specific, countable values (often integers). This lesson focuses on discrete random variables.

Representing Distributions

A probability distribution can be represented in three equivalent ways:

1. Table

x

123456

P (X = x)

\frac{1}{6}

\frac{1}{6}

\frac{1}{6}

\frac{1}{6}

\frac{1}{6}

\frac{1}{6}

2. PMF (Formula)

P (X = x) =

{

\frac{1}{6}

x = 1, 2, 3, 4, 5, 6

Validity Condition

For ANY valid probability distribution: $\sum P (X = x) = 1$ (all probabilities must sum to 1)

Probability Mass Functions

Definition

A probability mass function (PMF) is a formula that gives

P (X = x)

for each value of x. It must always specify the domain (which values of x are valid).

Simple PMF

P (X = x) =

{

\frac{x}{10}

x = 1, 2, 3, 4

One formula applies to all x values

Piecewise PMF

P (X = x) =

{

\frac{1}{8}

x = 0, 3

\frac{3}{8}

x = 1, 2

Different formulas for different x values

Piecewise Functions

When probabilities follow different patterns for different values, we use a piecewise function. This often occurs when deriving distributions from experiments.

Converting Between Forms

PMF → Table

Method: Substitute each x value into the formula

P (X = x) =

{

\frac{x}{10}

x = 1, 2, 3, 4

x

1234

P

0.10.20.30.4

Table → PMF

Method: Find a pattern relating x to P(X=x)

x

123

P

\frac{1}{14}

\frac{4}{14}

\frac{9}{14}

Pattern: 1, 4, 9 = 1², 2², 3²

P (X = x) =

{

\frac{x ^{2}}{14}

x = 1, 2, 3

Always Verify

After converting, check that

\sum P (X = x) = 1

. This confirms your formula is correct.

01Worked Example

Three fair coins are tossed. Let X = the number of heads.

(a) Write down all possible outcomes and the sample space of X.

(b) Write the probability distribution of X as: (i) a table (ii) a PMF.

(a)

Outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT (8 total)

•

Sample space: {0, 1, 2, 3}

(b)

Count outcomes for each X value:

•

X=0: TTT (1) → P=1/8; X=1: HTT,THT,TTH (3) → P=3/8

•

X=2: HHT,HTH,THH (3) → P=3/8; X=3: HHH (1) → P=1/8

(i)

Table: x = 0, 1, 2, 3 with P = 1/8, 3/8, 3/8, 1/8

(ii)

PMF:

P (X = x) =

\frac{3}{8}

x = 1, 2

02Worked Example

A biased four-sided dice has faces 1, 2, 3, 4. Given $P (X = x) = \frac{k}{x}$ for x = 1, 2, 3, 4:

(a) Find the value of k. (b) Give the distribution in table form. (c) Find P(X > 2).

(a)

Sum of probabilities = 1:

•

\frac{k}{1} + \frac{k}{2} + \frac{k}{3} + \frac{k}{4} = 1

•

k (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}) = k (\frac{12 + 6 + 4 + 3}{12}) = k \times \frac{25}{12} = 1

•

k = \frac{12}{25}

(b)

P(1)=12/25, P(2)=6/25, P(3)=4/25, P(4)=3/25

(c)

P (X > 2) = P (3) + P (4) = \frac{4}{25} + \frac{3}{25} = \frac{7}{25}

03Worked Example

A discrete random variable X has the piecewise PMF:

P (X = x) =

$kx$ $x = 1, 3$

k (x - 1)

x = 2, 4

(a) Find k. (b) Find P(X > 1).

(a)

Calculate each probability:

•

P(1) = k(1) = k; P(3) = k(3) = 3k

•

P(2) = k(2-1) = k; P(4) = k(4-1) = 3k

•

Sum = 1: k + k + 3k + 3k = 8k = 1 →

k = \frac{1}{8}

(b)

P(X > 1) = P(2) + P(3) + P(4) = 1/8 + 3/8 + 3/8 =

\frac{7}{8}

Finding Probabilities

$P (X \leq k)$ and $P (X < k)$

Add all probabilities where condition is satisfied

P (X \leq 3) = P (0) + P (1) + P (2) + P (3)

P (X < 3) = P (0) + P (1) + P (2)

$P (X > k)$ and $P (X \geq k)$

Use complement or add directly

P (X > k) = 1 - P (X \leq k)

P (X \geq k) = 1 - P (X < k)

Ranges: $P (a < X \leq b)$

Carefully check which endpoints are included:

P (1 < X \leq 4)

includes x = 2, 3, 4 (NOT 1)

P (1 \leq X < 4)

includes x = 1, 2, 3 (NOT 4)

Watch Out

If asked for P(X > 4) but the sample space only goes up to 4, then P(X > 4) = 0.

04Worked Example

X has the distribution shown:

x

0123

P (X = x)

\frac{1}{8}

\frac{1}{4}

\frac{1}{2}

\frac{1}{8}

Find: (a)

P (1 < X \leq 3)

(b)

P (X < 2)

(c)

P (X > 3)

(a)

P (1 < X \leq 3)

includes x = 2, 3 (NOT 1)

•

= P (2) + P (3) = \frac{1}{2} + \frac{1}{8} = \frac{5}{8}

(b)

P (X < 2)

includes x = 0, 1

•

= P (0) + P (1) = \frac{1}{8} + \frac{1}{4} = \frac{3}{8}

(c)

P (X > 3) = 0

(no values greater than 3 in sample space)

Verifying a Valid PMF

Two Conditions

1. All probabilities non-negative:

P (X = x) \geq 0

for all x

2. Probabilities sum to 1:

\sum P (X = x) = 1

Valid PMF ✓

P (X = x) =

{

\frac{x}{6}

x = 1, 2, 3

Sum = 1/6 + 2/6 + 3/6 = 1 ✓

Invalid PMF ✗

P (X = x) =

{

\frac{2}{x ^{2}}

x = 2, 3, 4

Sum = 2/4 + 2/9 + 2/16 ≈ 0.847 ≠ 1 ✗

Exam Tip

If asked "Explain why this is not a valid probability distribution", calculate the sum and show it ≠ 1.

05Worked Example

Marie says X has PMF:

P (X = x) =

$\frac{2}{x^2}$

x = 2, 3, 4

(a) Explain why this is not a valid probability distribution.

(b) Given the correct PMF is

P (X = x) = \frac{k}{x ^{2}}

for x = 2, 3, 4, find k.

(a)

Sum of probabilities:

•

\frac{2}{4} + \frac{2}{9} + \frac{2}{16} = \frac{1}{2} + \frac{2}{9} + \frac{1}{8} = \frac{36 + 16 + 9}{72} = \frac{61}{72} \neq = 1

•

Sum ≠ 1, so not a valid probability distribution.

(b)

\frac{k}{4} + \frac{k}{9} + \frac{k}{16} = 1

•

k (\frac{36 + 16 + 9}{144}) = 1

→

k \times \frac{61}{144} = 1

→

k = \frac{144}{61}

Discrete Uniform Distribution

Definition

A discrete uniform distribution is one where all outcomes are equally likely. If there are n outcomes, each has probability

\frac{1}{n}

.

Example: Fair Die

n = 6 outcomes

P (X = x) =

{

\frac{1}{6}

x = 1, 2, ..., 6

General Form

P (X = x) =

{

\frac{1}{n}

all

x

in sample space

Identifying Uniform

Look for: "fair" dice/spinners, "random" selection, "equally likely" outcomes.

06Worked Example

X takes integer values 1 to 50 with discrete uniform distribution. Find:

(a) P(X = 1) (b) P(X ≥ 28) (c) P(13 < X < 42)

(a)

Uniform with 50 values:

P (X = 1) = \frac{1}{50}

(b)

X ≥ 28 means X = 28, 29, ..., 50 (count: 50 - 28 + 1 = 23 values)

•

P (X \geq 28) = 23 \times \frac{1}{50} = \frac{23}{50}

(c)

13 < X < 42 means X = 14, 15, ..., 41 (count: 41 - 14 + 1 = 28 values)

•

P (13 < X < 42) = 28 \times \frac{1}{50} = \frac{28}{50} = \frac{14}{25}

Deriving Distributions

Sometimes you need to derive a probability distribution from an experiment's stopping rule:

"Until" Problems

When an experiment continues "until" something happens (or reaches a maximum), work out each probability systematically.

Method

Identify what the random variable represents

List all possible values (sample space)

For each value, work out how that outcome occurs

Calculate probabilities using multiplication (independent events)

For the final value, use: P(last) = 1 - P(all earlier values)

Worked Example07

A spinner has P(Red) = 2/5 and P(Blue) = 3/5. It is spun until red OR it's been spun 4 times. Let S = number of spins. Find the distribution of S.

1

Sample space: S = 1, 2, 3, or 4

2

P(S=1) = P(Red first spin) =

\frac{2}{5}

3

P(S=2) = P(Blue then Red) =

\frac{3}{5} \times \frac{2}{5} = \frac{6}{25}

4

P(S=3) = P(B, B, R) =

\frac{3}{5} \times \frac{3}{5} \times \frac{2}{5} = \frac{18}{125}

5

P(S=4) = 1 - (2/5 + 6/25 + 18/125) = 1 - 50/125 - 30/125 - 18/125 =

\frac{27}{125}

Table: S = 1, 2, 3, 4 with P = 2/5, 6/25, 18/125, 27/125

Modelling Real Situations

Assumptions

Independence of trials

Constant probabilities

Outcomes mutually exclusive

Random selection

Limitations

Real situations rarely perfectly random

Probabilities may change

Events may not be independent

Model simplifies reality

Watch Out

A model only approximates real behaviour. Real experiments will never exactly match the model, but it provides a useful way of analysing outcomes.

Practice

Question 1:

P (X = x) =

$0.1$

x = - 2, - 1

$\beta$

x = 0, 1

$0.2$

x = 2

(a) Find β. (b) Construct a table for the distribution. (c) Find P(−1 ≤ X < 2).

(a)

0.1 + 0.1 + β + β + 0.2 = 1 → 2β = 0.6 →

β = 0.3

(b)

x: −2, −1, 0, 1, 2 → P: 0.1, 0.1, 0.3, 0.3, 0.2

(c)

P(−1 ≤ X < 2) = P(−1) + P(0) + P(1) = 0.1 + 0.3 + 0.3 =

0.7

Practice

Question 2:

x

1234

P

\frac{1}{3}

\frac{1}{3}

k

\frac{1}{4}

(a) Find k. (b) Write a PMF for X.

(a)

1/3 + 1/3 + k + 1/4 = 1 → k = 1 − 8/12 − 3/12 =

\frac{1}{12}

(b)

No simple pattern; piecewise PMF:

P (X = x) =

x = 1, 2

x = 3

x = 4

Practice

Question 3:

(a) Write down the distribution of S in table form. (b) Find P(S > 2).

(a)

P(S=1) = 2/3; P(S=2) = 1/3 × 2/3 = 2/9; P(S=3) = 1/3 × 1/3 × 2/3 = 2/27

•

P(S=4) = 1 − (2/3 + 2/9 + 2/27) = 1 − 26/27 = 1/27

•

Table: S = 1, 2, 3, 4 with P = 2/3, 2/9, 2/27, 1/27

(b)

P(S > 2) = P(3) + P(4) = 2/27 + 1/27 =

\frac{3}{27} = \frac{1}{9}

Summary

Key Concepts

ConceptKey Points

Random VariableValue depends on random outcome; sample space = all possible values

RepresentationsTable, PMF (formula), or vertical line diagram - all equivalent

PMF TypesSimple (one formula) or piecewise (different rules for different x)

ValidityAll P ≥ 0 AND ΣP(X = x) = 1

UniformAll outcomes equally likely: P(X = x) = 1/n

Exam Skills

Convert between table, PMF, and diagram forms

Use ΣP = 1 to find unknown constants

Derive distributions from "until" stopping rules

Verify whether a function is a valid PMF