Interval probabilities by integration | Year 12

01 / LEARN

The idea, step by step

Translate the event first

Sketch the support and mark the event before integrating. Only the part inside the support can contribute probability. For a continuous variable, strict and inclusive endpoints give the same answer when they differ at just one value.

For a probability between a and b, integrate in increasing order. A reversed signed integral could be negative, but a probability cannot. Keep exact fractions during the calculation and round once at the end.

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

Choose a convenient area

An upper tail can be found directly or as one minus the lower area. Add areas for disjoint intervals but do not include a gap between them. Check that the final answer is between zero and one.

A conditional question changes the reference group. Divide the intersection area by the area of the conditioning event. Use the same density in both integrals, and check that the conditioning event has positive probability.

A common trap

Using a density height as a tail probability.

Instead: Integrate over the event; substituting an endpoint only gives a height.

02 / WORKED EXAMPLES

See the thinking

Example 1

A middle interval

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

$P = \int_{0.2}^{0.6} 2 x d x$ .
$P = {0.6}^{2} - {0.2}^{2} = 0.32$ .

0.32.

Example 2

An upper tail

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

The lower area is ${[x^{3}]}_{0}^{0.5} = 0.125$ .
Upper area $= 1 - 0.125 = 0.875$ .

0.875.

Check your understanding

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

Show answer

0.25.

${[x^{2}]}_{0}^{0.5} = 0.25$ .

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF 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 $P (1 / 4 < X < 3 / 4)$ .
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 1/12 and the event width is 5.
$5 / 12$ .
For density $6 x (1 - x)$ on [0,1], find $P (1 / 4 < X < 3 / 4)$ .
3 marks
Use $3 x^{2} - 2 x^{3}$ as an antiderivative.
Its endpoint difference is $27 / 32 - 5 / 32 = 11 / 16$ .
$11 / 16$ .
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.
$64 / 91$ .
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 ↗

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