Interval probabilities by integration | Year 12

01 / LEARN

Understand the idea.

Translate the event first

Translate the event into an interval, then integrate the density over that interval. Intersect the event with the support first: any part outside the support contributes zero. For density models, changing an endpoint from strict to inclusive changes no probability.

Integrate from the lower bound to the upper bound. Reversing limits changes an integral's sign, but a probability cannot be negative. Use an antiderivative to subtract the lower endpoint value from the upper endpoint value, retaining exact values until the final rounding.

P (a < X < b) = \int_{a}^{b} f (x) d x

Choose a convenient area

Choose the simplest equivalent area. An upper tail may be found directly or by subtracting the lower area from $1$ . For disjoint intervals, add their separate areas without accidentally including the gaps between them. Check the final result lies in $[0, 1]$ .

For conditional probability, the conditioning event becomes the reference region. Calculate the intersection area and divide by the conditioning event's area, which must be positive. Dividing adjusts the scale; it does not change the original density used in those integrals.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A middle interval

For density $2 x$ on [0,1], find $P (0.2 < X < 0.6)$ .

Worked solution

1
Set the integral over the required interval.
$P (0.2 < X < 0.6) = \int_{0.2}^{0.6} 2 x d x$
Both event boundaries lie inside the support from zero to one.
2
Use an antiderivative.
$\int_{0.2}^{0.6} 2 x d x = {[x^{2}]}_{0.2}^{0.6}$
Differentiating the square returns the density.
3
Subtract the endpoint values.
$P (0.2 < X < 0.6) = {0.6}^{2} - {0.2}^{2} = 0.36 - 0.04 = 0.32$
The difference measures area between the two boundaries rather than a density height.

Answer

0.32.

EXAMPLE 02

An upper tail

For density $3 x^{2}$ on [0,1], find $P (X > 0.5)$ .

Worked solution

1
Choose the complementary lower interval.
$P (X > 0.5) = 1 - P (X \leq 0.5)$
The density has total probability one and a single endpoint contributes no mass.
2
Integrate the lower area.
$P (X \leq 0.5) = \int_{0}^{0.5} 3 x^{2} d x = {[x^{3}]}_{0}^{0.5} = 0.125$
The lower endpoint of the support is zero.
3
Calculate the upper tail.
$P (X > 0.5) = 1 - 0.125 = 0.875$
This retains exactly the area from one half to one.

Answer

0.875.

03 / YOUR TURN

Try it yourself.

Check your understanding

For density $2 x$ on [0,1], find $P (X < 0.5)$ .

Show answer & reasoning

0.25.

Use the area from $0$ to $0.5$ : $P (X < 0.5) = \int_{0}^{0.5} 2 x d x = {[x^{2}]}_{0}^{0.5} = 0.25$ .

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

For density $2 x$ on $[0, 1]$ , find $P (0.3 < X < 0.9)$ .
2 marks
For density $3 x^{2}$ on $[0, 1]$ , find $P (X < 0.2)$ .
2 marks
A waiting time is uniform on $[0, 12]$ minutes. Find the chance of waiting between 3 and 8 minutes.
2 marks
For density $6 x (1 - x)$ on $[0, 1]$ , find $P (\frac{1}{4} < X < \frac{3}{4})$ .
3 marks
For density $2 x$ on $[0, 1]$ , find $P (X > 0.6 ∣ X > 0.3)$ .
3 marks
Explain why including either endpoint of a continuous interval does not change its probability.
2 marks

Read the worked solutions

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

For density $2 x$ on $[0, 1]$ , find $P (0.3 < X < 0.9)$ .
2 marks
${0.9}^{2} - {0.3}^{2} = 0.72$ .
0.72.
For density $3 x^{2}$ on $[0, 1]$ , find $P (X < 0.2)$ .
2 marks
${[x^{3}]}_{0}^{0.2} = {0.2}^{3}$ .
0.008.
A waiting time is uniform on $[0, 12]$ minutes. Find the chance of waiting between 3 and 8 minutes.
2 marks
The density is $\frac{1}{12}$ and the event width is 5.
$\frac{5}{12}$ .
For density $6 x (1 - x)$ on $[0, 1]$ , find $P (\frac{1}{4} < X < \frac{3}{4})$ .
3 marks
Use $3 x^{2} - 2 x^{3}$ as an antiderivative.
Its endpoint difference is $\frac{27}{32} - \frac{5}{32} = \frac{11}{16}$ .
$\frac{11}{16}$ .
For density $2 x$ on $[0, 1]$ , find $P (X > 0.6 ∣ X > 0.3)$ .
3 marks
The numerator is $1 - {0.6}^{2} = 0.64$ .
The denominator is $1 - {0.3}^{2} = 0.91$ ; divide.
$\frac{64}{91}$ .
Explain why including either endpoint of a continuous interval does not change its probability.
2 marks
A finite collection of singleton values adds no area.
Each endpoint has probability zero.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods: continuous random variables, normal distributions and sampling distributions of sample means. Original lessons and practice.

Official curriculum source ↗

In this lesson

Understand the idea.

Translate the event first

Choose a convenient area

See the method in action.

A middle interval

Set the integral over the required interval.

Use an antiderivative.

Subtract the endpoint values.

An upper tail

Choose the complementary lower interval.

Integrate the lower area.

Calculate the upper tail.

Try it yourself.

Check your understanding

Keep practising