Probability functions and cumulative probabilities | Year 12

01 / LEARN

Understand the idea.

Mass at a point, total to the left

The probability mass function $p (x) = P (X = x)$ gives the probability at one value. The cumulative distribution function $F (x) = P (X \leq x)$ adds all mass at or below its argument, including an endpoint that is possible.

Translate strict inequalities into actual values before adding. For an integer-valued variable, $X < 3$ means $X \leq 2$ . If the support is not consecutive integers, list its values rather than assuming every integer can occur.

Recover probabilities from the jumps

A discrete cumulative function is constant between support values and jumps at each value with positive probability. The jump is that value's probability. Therefore a CDF cannot decrease, and its far-left and far-right limits are $0$ and $1$ .

Subtract cumulative totals to isolate an interval. For an integer-valued variable $X$ and integer endpoints $a, b$ , $P (a \leq X \leq b) = F (b) - F (a - 1)$ . The subtraction removes only values below the lower endpoint. For other supports, use the cumulative total immediately before that endpoint.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Build and use cumulative values

X has probabilities 0.1, 0.3, 0.4, 0.2 at 0, 1, 2, 3. Find F(1), P(X > 1) and P(1 < X ≤ 3).

Probability distribution of X
x	P(X = x)
0	1/10
1	3/10
2	2/5
3	1/5

Worked solution

1
Accumulate through one.
$F (1) = P (X = 0) + P (X = 1) = 0.1 + 0.3 = 0.4$
The definition of $F (1)$ includes both possible values no larger than $1$ .
2
Take the complementary upper event.
$P (X > 1) = 1 - F (1) = 1 - 0.4 = 0.6$
The events $X \leq 1$ and $X > 1$ partition all possible values.
3
Identify the values in the final interval.
$P (1 < X \leq 3) = P (X = 2) + P (X = 3) = 0.4 + 0.2 = 0.6$
The lower endpoint is excluded, while the upper endpoint includes the remaining support value $3$ .

Answer

F(1) = 0.4; both event probabilities are 0.6.

EXAMPLE 02

Read a jump

An integer-valued variable has F(0) = 0.2, F(1) = 0.55 and F(2) = 1. Find P(X = 1) and P(X ≥ 1).

Worked solution

1
Subtract the cumulative totals around one.
$P (X = 1) = F (1) - F (0) = 0.55 - 0.20 = 0.35$
For an integer-valued variable there are no possible values strictly between zero and one.
2
Write the complement of at least one.
${X \geq 1}^{c} = {X \leq 0}$
The complementary event must exclude one rather than remove it.
3
Calculate the upper probability.
$P (X \geq 1) = 1 - F (0) = 1 - 0.2 = 0.8$
The given cumulative value already combines every outcome below one.

Answer

P(X = 1) = 0.35; P(X ≥ 1) = 0.8.

03 / YOUR TURN

Try it yourself.

Check your understanding

If F(2) = 0.7, what is P(X > 2)?

Show answer & reasoning

0.3.

Since $F (2) = P (X \leq 2) = 0.7$ , its complement is $P (X > 2) = 1 - 0.7 = 0.3$ .

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

For probabilities 0.2, 0.5 and 0.3 at $x = 0$ , 1, 2, give $F (0)$ , $F (1)$ and $F (2)$ .
3 marks
For the same distribution, find $P (X < 2)$ .
2 marks
An integer-valued $X$ has $F (2) = 0.65$ and $F (3) = 0.90$ . Find $P (X = 3)$ .
2 marks
$X$ takes only the values 2, 5 and 8, with probabilities 0.1, 0.6 and 0.3. Find $F (6)$ .
2 marks
For integer-valued $X$ , $F (1) = 0.3$ and $F (4) = 0.95$ . Find $P (2 \leq X \leq 4)$ .
2 marks
A proposed cumulative function has $F (1) = 0.7$ and $F (2) = 0.6$ . Explain why it is impossible.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

For probabilities 0.2, 0.5 and 0.3 at $x = 0$ , 1, 2, give $F (0)$ , $F (1)$ and $F (2)$ .
3 marks
Take running totals: 0.2; $0.2 + 0.5$ ; $0.2 + 0.5 + 0.3$ .
0.2, 0.7, 1.
For the same distribution, find $P (X < 2)$ .
2 marks
The event includes $x = 0$ and $x = 1$ only.
Add $0.2 + 0.5$ .
0.7.
An integer-valued $X$ has $F (2) = 0.65$ and $F (3) = 0.90$ . Find $P (X = 3)$ .
2 marks
Subtract the cumulative total before 3 from the total at 3.
$0.90 - 0.65 = 0.25$ .
0.25.
$X$ takes only the values 2, 5 and 8, with probabilities 0.1, 0.6 and 0.3. Find $F (6)$ .
2 marks
The values no larger than 6 are 2 and 5.
Add $0.1 + 0.6$ ; there need not be mass at 6.
0.7.
For integer-valued $X$ , $F (1) = 0.3$ and $F (4) = 0.95$ . Find $P (2 \leq X \leq 4)$ .
2 marks
$F (4)$ includes all values up to 4.
Subtract $F (1)$ , which excludes everything below 2: $0.95 - 0.3$ .
0.65.
A proposed cumulative function has $F (1) = 0.7$ and $F (2) = 0.6$ . Explain why it is impossible.
2 marks
The event $X \leq 1$ is contained in $X \leq 2$ .
Adding possible values cannot reduce total probability.
A cumulative probability cannot decrease.

Checking saved progress…

Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods: discrete random variables and probability distributions, expected value and variance, transformations and independent sums, Bernoulli and binomial models. Original supporting lessons and exercises follow the approved Ascent chapter structure.

Official curriculum source ↗

In this lesson

Understand the idea.

Mass at a point, total to the left

Recover probabilities from the jumps

See the method in action.

Build and use cumulative values

Accumulate through one.

Take the complementary upper event.

Identify the values in the final interval.

Read a jump

Subtract the cumulative totals around one.

Write the complement of at least one.

Calculate the upper probability.

Try it yourself.

Check your understanding

Keep practising