Statistical Distributions

What you'll learn

How to describe a discrete random variable using a probability distribution.
How to build and use a cumulative distribution function.
How to recognise discrete uniform and binomial distributions.
How to handle common exam-style tasks: missing probabilities, inequalities, independence, transformations, and expected counts.

1. Discrete random variables and probability distributions

A random variable is a variable whose value depends on chance.

A random variable is discrete if it can only take separate, countable values, such as 0, 1, 2, 3 or the score on a spinner.

Definition

Discrete probability distribution

A discrete probability distribution lists every possible value of a random variable and the probability of each value. The probabilities must:

each be between 0 and 1
add up to 1

For example, if $X$ is the number scored on a spinner, then $P (X = x)$ means “the probability that $X$ takes the value $x$ ”.

Key Idea

The total probability rule

For any complete probability distribution:

∑P(X=x)=1\sum P(X=x)=1

This is usually the quickest way to find a missing constant.

Worked example: finding a missing probability

A discrete random variable $X$ has this probability distribution:

A probability bar chart for the distribution, highlighting that only the values 2 and 3 are included in P(X>1.4).

Value of X	0	1	2	3
Probability	0.18	a	0.27	0.35

Find the value of $a$ , then find $P (X > 1.4)$ .

Example

Missing value and an inequality

Use the fact that all probabilities add to 1:
$0.18 + a + 0.27 + 0.35 = 1$
Solve for $a$ :
$a = 1 - 0.18 - 0.27 - 0.35 = 0.20$
Interpret the inequality carefully. Since $X$ can only be 0, 1, 2 or 3, the values greater than 1.4 are 2 and 3.
Add the relevant probabilities:
$P (X > 1.4) = P (X = 2) + P (X = 3) = 0.27 + 0.35 = 0.62$

Common Mistake

Treating a discrete variable as continuous

If $X$ only takes integer values, then $P (X > 1.4)$ is the same as $P(X≥2)P(X\ge 2)$ . Do not try to include “part” of the probability at 1 or 2.

2. Cumulative distribution functions

The cumulative distribution function, often written as $F (x)$ , gives the probability that the random variable is less than or equal to a value.

Definition

Cumulative distribution function

For a random variable $X$ ,

F(x)=P(X≤x)F(x)=P(X\le x)

So $F (x)$ is the running total of probabilities up to and including $x$ .

A cumulative distribution for a discrete variable increases in jumps, because probability is only added at the possible values of $X$ .

A discrete probability distribution and its cumulative step graph, showing how the running total increases only at possible values of X.

Discrete distribution bars and cumulative step graph

Worked example: constructing a cumulative table

A random variable $X$ has the distribution below.

Value of X	1	2	3	4
Probability	0.10	0.25	0.40	0.25

Construct a table for $F (x)$ .

Example

Building the cumulative distribution

Start with the first value:
$F(1)=P(X≤1)=0.10F(1)=P(X\le 1)=0.10$
Add the probability at 2:
$F(2)=P(X≤2)=0.10+0.25=0.35F(2)=P(X\le 2)=0.10+0.25=0.35$
Add the probability at 3:
$F(3)=P(X≤3)=0.10+0.25+0.40=0.75F(3)=P(X\le 3)=0.10+0.25+0.40=0.75$
Add the probability at 4:
$F (4) = 1$
Write the cumulative table:

Value of X	1	2	3	4
Cumulative probability	0.10	0.35	0.75	1.00

Tip

CDF sanity checks

A cumulative distribution should never go down, and the final value should be 1.

3. Discrete uniform distributions

A discrete uniform distribution occurs when each possible outcome is equally likely.

Definition

Discrete uniform distribution

If a random variable has $n$ equally likely values, then each value has probability $1n\frac{1}{n}$ .

Common examples include a fair die, a fair spinner with equal sectors, or choosing one card from equally likely labelled cards.

Worked example: fair spinner

A fair spinner has four equal sectors labelled 1, 2, 3 and 4. Let $Y$ be the number landed on.

A fair four-sector spinner illustrates that each outcome has equal probability in a discrete uniform distribution.

Example

Writing down a uniform distribution

Since the spinner is fair, all four outcomes are equally likely.
There are four possible values, so each probability is:
$14\frac{1}{4}$
The probability distribution is:

Value of Y	1	2	3	4
Probability	1/4	1/4	1/4	1/4

The name of the distribution is the discrete uniform distribution.

4. The binomial distribution

The binomial distribution models the number of successes in a fixed number of repeated trials.

A binomial model is built from repeated independent trials, each with success probability p and failure probability 1-p.

Definition

Binomial distribution

We write $X∼B(n,p)X\sim B(n,p)$ when:

there are $n$ trials
each trial has two outcomes: success or failure
the probability of success is constant, $p$
the trials are independent
$X$ counts the number of successes

The formula for exactly $r$ successes is:

P(X=r)=(nr)pr(1−p)n−rP(X=r)=\binom{n}{r}p^r(1-p)^{n-r}

Here, $(nr)\binom{n}{r}$ counts how many different ways the successes can be arranged.

Worked example: exactly, less than, and at least

A student has probability 0.3 of answering a question correctly. They answer 12 independent questions. Let $X$ be the number answered correctly.

$A binomial probability bar chart for X\sim B(12,0.3) helps visualise exactly 4, less than 3, and at least 5 correct answers.$

Find $P (X = 4)$ , $P (X < 3)$ , and $P(X≥5)P(X\ge 5)$ .

Example

Using a binomial model

Identify the distribution:
$X∼B(12,0.3)X\sim B(12,0.3)$
For exactly 4 correct answers, use the binomial formula:
$P(X=4)=(124)(0.3)4(0.7)8P(X=4)=\binom{12}{4}(0.3)^4(0.7)^8$
Evaluate:
$P(X=4)≈0.231P(X=4)\approx 0.231$
For $P (X < 3)$ , include 0, 1 and 2:
$P(X<3)=P(X≤2)P(X<3)=P(X\le 2)$
Using a calculator’s binomial cumulative function, or by adding the three terms:
$P(X≤2)≈0.253P(X\le 2)\approx 0.253$
For $P(X≥5)P(X\ge 5)$ , use the complement:
$P(X≥5)=1−P(X≤4)P(X\ge 5)=1-P(X\le 4)$
Evaluate:
$P(X≥5)≈1−0.7237=0.276P(X\ge 5)\approx 1-0.7237=0.276$

Common Mistake

Wrong boundary in binomial inequalities

For integer-valued $X$ , $P (X < 3)$ means $P(X≤2)P(X\le 2)$ , not $P(X≤3)P(X\le 3)$ . Similarly, $P (X > 4)$ means $P(X≥5)P(X\ge 5)$ .

5. Checking binomial assumptions

Before using a binomial model, check that the situation really matches the conditions.

For example, “number of hits in 8 darts” can be binomial if each throw has the same probability of a hit and throws are independent.

But “number of goals in a football match” might be less realistic as binomial, because scoring chances are not necessarily independent and the probability of scoring may change during the match.

Worked example: modelling hits

A player throws 6 balls at a bucket. The probability of a successful throw is 0.4. Assume throws are independent.

Let $X$ be the number of successful throws.

Find $P (X = 2)$ . Then find the probability that the first success is on the fourth throw.

A sequence diagram distinguishes exactly two successes in six throws from the event that the first success occurs on throw 4.

Example

Binomial probability and first success

For the number of successes in 6 throws:
$X∼B(6,0.4)X\sim B(6,0.4)$
Find exactly 2 successes:
$P(X=2)=(62)(0.4)2(0.6)4P(X=2)=\binom{6}{2}(0.4)^2(0.6)^4$
Evaluate:
$P(X=2)≈0.311P(X=2)\approx 0.311$
For the first success on the fourth throw, the first three throws must be failures and the fourth must be a success.
Multiply the probabilities:
$4)=(0.6)3(0.4)=0.0864P(\text{first success on throw }4)=(0.6)^3(0.4)=0.0864$

Common Mistake

First success is not the same as exactly one success

“First success on throw 4” says what happens on throws 1 to 4 only: fail, fail, fail, success. It does not automatically describe later throws unless the question says so.

6. Expected counts from a probability

If an event has probability $p$ and it is repeated across $N$ independent occasions, the expected number of occasions on which it happens is:

N p

This is useful when a binomial model is used inside a larger context.

Worked example: expected number of matches

A team’s number of goals in a match is modelled by $X∼B(9,0.2)X\sim B(9,0.2)$ .

Using this model, find the expected number of matches with no goals in a 30-match season.

Example

Expected number using a model

First find the probability of no goals in one match:
$P(X=0)=(90)(0.2)0(0.8)9P(X=0)=\binom{9}{0}(0.2)^0(0.8)^9$
Simplify:
$P(X=0)=(0.8)9≈0.134P(X=0)=(0.8)^9\approx 0.134$
Multiply by the number of matches:
$30×0.134≈4.0330\times 0.134\approx 4.03$
So the expected number of goalless matches is about 4.

Tip

Interpreting expected values

An expected number does not have to be a whole number. It is a long-run average, not a guaranteed outcome.

7. Combining independent random variables

Sometimes a question involves two independent attempts, games, or random variables.

If two events are independent, the result of one does not affect the result of the other. For independent events, you multiply probabilities.

Worked example: total score from two games

In one game, a player can score 0, 5, 10 or 15 points with probabilities:

Score	0	5	10	15
Probability	0.25	0.30	0.20	0.25

The player plays twice independently. Find the probability that the total score is 20.

A two-way grid shows all score pairs from two independent games and highlights the pairs that total 20.

Example

Adding scores from independent games

List the score pairs that total 20:
$(15,5)(5,15),\ (10,10),\ (15,5)$
Multiply probabilities within each pair because the games are independent:
$P(5,15)=0.30×0.25P(5,15)=0.30\times 0.25$
Do the same for all possible pairs and add them:
$20)=(0.30)(0.25)+(0.20)(0.20)+(0.25)(0.30)P(\text{total }20)=(0.30)(0.25)+(0.20)(0.20)+(0.25)(0.30)$
Evaluate:
$20)=0.075+0.040+0.075=0.190P(\text{total }20)=0.075+0.040+0.075=0.190$

8. Transforming a random variable

A transformed random variable is made from another random variable using a formula, such as $Y = 7 - 2 X$ .

To find probabilities involving $Y$ , replace $Y$ with its formula and solve the inequality for $X$ .

Worked example: using a transformation

Let $X$ take the values 1, 2, 3 and 4. Suppose $Y = 9 - 3 X$ . Find $P(Y+X≥2)P(Y+X\ge 2)$ .

A mapping table for Y=9-3X shows which discrete X values satisfy the transformed inequality.

Example

Rewriting the inequality in terms of X

Substitute $Y = 9 - 3 X$ into the inequality:
$Y+X≥2Y+X\ge 2$
Simplify in terms of $X$ :
$9−3X+X≥29-3X+X\ge 2$
Solve:
$9−2X≥29-2X\ge 2$
Continue solving:
$−2X≥−7-2X\ge -7$
Divide by a negative number, remembering to reverse the inequality:
$X≤3.5X\le 3.5$
Since $X$ can only be 1, 2, 3 or 4, this means $X = 1, 2, 3$ .
Add the relevant probabilities from the distribution of $X$ .

Common Mistake

Forgetting to reverse the inequality

When you divide an inequality by a negative number, the inequality sign changes direction.

Exam technique

In the exam

Check whether the variable is discrete: inequalities usually need careful conversion, such as $X < 4$ becoming $X≤3X\le 3$ .
For a probability table, always use “probabilities add to 1” first if a constant is missing.
For binomial questions, write $X∼B(n,p)X\sim B(n,p)$ before calculating so you are clear about the number of trials and success probability.
Use complements for phrases like “more than”, “at least”, and “at least one” when it saves time.
When combining independent events, multiply along a specific route and add different possible routes.

Self review

Check yourself

Can you explain the difference between $P (X = x)$ and $F (x)$ ?
If $X∼B(20,0.15)X\sim B(20,0.15)$ , what values of $n$ and $p$ are being used?
For a discrete random variable, why is $P (X < 6)$ usually rewritten as $P(X≤5)P(X\le 5)$ ?

Statistical Distributions

What you'll learn

How to describe a discrete random variable using a probability distribution.
How to build and use a cumulative distribution function.
How to recognise discrete uniform and binomial distributions.
How to handle common exam-style tasks: missing probabilities, inequalities, independence, transformations, and expected counts.

1. Discrete random variables and probability distributions

A random variable is a variable whose value depends on chance.

A random variable is discrete if it can only take separate, countable values, such as 0, 1, 2, 3 or the score on a spinner.

Definition

Discrete probability distribution

A discrete probability distribution lists every possible value of a random variable and the probability of each value. The probabilities must:

each be between 0 and 1
add up to 1

For example, if $X$ is the number scored on a spinner, then $P (X = x)$ means “the probability that $X$ takes the value $x$ ”.

Key Idea

The total probability rule

For any complete probability distribution:

∑P(X=x)=1\sum P(X=x)=1

This is usually the quickest way to find a missing constant.

Worked example: finding a missing probability

A discrete random variable $X$ has this probability distribution:

A probability bar chart for the distribution, highlighting that only the values 2 and 3 are included in P(X>1.4).

Value of X	0	1	2	3
Probability	0.18	a	0.27	0.35

Find the value of $a$ , then find $P (X > 1.4)$ .

Example

Missing value and an inequality

Use the fact that all probabilities add to 1:
$0.18 + a + 0.27 + 0.35 = 1$
Solve for $a$ :
$a = 1 - 0.18 - 0.27 - 0.35 = 0.20$
Interpret the inequality carefully. Since $X$ can only be 0, 1, 2 or 3, the values greater than 1.4 are 2 and 3.
Add the relevant probabilities:
$P (X > 1.4) = P (X = 2) + P (X = 3) = 0.27 + 0.35 = 0.62$

Common Mistake

Treating a discrete variable as continuous

If $X$ only takes integer values, then $P (X > 1.4)$ is the same as $P(X≥2)P(X\ge 2)$ . Do not try to include “part” of the probability at 1 or 2.

2. Cumulative distribution functions

The cumulative distribution function, often written as $F (x)$ , gives the probability that the random variable is less than or equal to a value.

Definition

Cumulative distribution function

For a random variable $X$ ,

F(x)=P(X≤x)F(x)=P(X\le x)

So $F (x)$ is the running total of probabilities up to and including $x$ .

A cumulative distribution for a discrete variable increases in jumps, because probability is only added at the possible values of $X$ .

A discrete probability distribution and its cumulative step graph, showing how the running total increases only at possible values of X.

Discrete distribution bars and cumulative step graph

Worked example: constructing a cumulative table

A random variable $X$ has the distribution below.

Value of X	1	2	3	4
Probability	0.10	0.25	0.40	0.25

Construct a table for $F (x)$ .

Example

Building the cumulative distribution

Start with the first value:
$F(1)=P(X≤1)=0.10F(1)=P(X\le 1)=0.10$
Add the probability at 2:
$F(2)=P(X≤2)=0.10+0.25=0.35F(2)=P(X\le 2)=0.10+0.25=0.35$
Add the probability at 3:
$F(3)=P(X≤3)=0.10+0.25+0.40=0.75F(3)=P(X\le 3)=0.10+0.25+0.40=0.75$
Add the probability at 4:
$F (4) = 1$
Write the cumulative table:

Value of X	1	2	3	4
Cumulative probability	0.10	0.35	0.75	1.00

Tip

CDF sanity checks

A cumulative distribution should never go down, and the final value should be 1.