A-level · Beta

Conditional probability

Highlight a two-way table to choose the right denominator and test independence.

P(B\mid A)=\frac{20}{50}\approx 0.4

Selected probability: 0.4
Numerator: both A and B: 20
Denominator: A: 50
P(A) × P(B): 0.2

P(B | A) uses 50 equally likely observations in its denominator.

Synthetic frequencies — each observation belongs to one cell
Event	B	Not B	Total
A	20	30	50
Not A	20	30	50
Total	40	60	100

Blue cell: numerator Tinted cells plus blue: denominator

Are A and B independent?

Yes in this constructed model: P(A ∩ B) = P(A)P(B).

$P(A\cap B)=0.2,\quad P(A)P(B)=0.2$

A and B are events, not people. All observations are equally likely when sampled from this table. These are synthetic counts; matching frequencies in a real sample alone would not establish population independence. Conditioning changes the sample space. Rounded probabilities are shown to 5 decimal places.

Predict, test and explain

Step 1 of 3

Predict

To calculate P(A | B) from a two-way frequency table, which total belongs in the denominator?

Explore and compare

Choose the investigation above. Predict what will change before you move a control.

Predict

Try it in the simulation. Change one thing at a time.

Explain

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Hints, self-check and connections

Try a self-check

Fixed values, separate from the diagram controls.

To calculate P(A | B) from a two-way frequency table, which total belongs in the denominator?

$P(A\mid B)=\frac{30}{30+20}$

Hint

Read the event after the vertical bar first: it tells you which group is now the whole sample space.

Reveal answer

Given B restricts the possible selections to the B group. Count A and B together, then divide by the B total. P(A | B) and P(B | A) can differ because their denominators differ.

$P(A\mid B)=0.6,\quad P(B\mid A)=0.75$

Build an explanation

Hint 1

Read the event after the vertical bar first: it tells you which group is now the whole sample space.

Hint 2

With a in both events, b in A only, c in B only and d in neither, P(A | B) = a/(a + c).

Hint 3

For a nonempty table, independence requires P(A ∩ B) = P(A)P(B). A conditional comparison can only be used when its denominator is nonzero.

Worked example

This example uses fixed values, separate from the diagram controls.

A table has 30 in A and B, 10 in A only, 20 in B only and 40 in neither. Find P(A | B) and P(B | A).

The B total is 30 + 20 = 50, while the A total is 30 + 10 = 40.
Given B, 30 of the 50 members are also in A: P(A | B) = 30/50 = 0.6.
Given A, 30 of the 40 members are also in B: P(B | A) = 30/40 = 0.75.

Answer: P(A | B) = 0.6 and P(B | A) = 0.75; reversing the condition changes the denominator.

Connect this idea

Watch out: Given B restricts the possible selections to the B group. Count A and B together, then divide by the B total. P(A | B) and P(B | A) can differ because their denominators differ. The frequencies describe a constructed table, with each member equally likely to be selected. A conditional probability is undefined when its conditioning group is empty. Independence here describes this table; it does not establish independence in a wider population or a causal relationship.

Specification and learning route

AQA M1, M2 · Edexcel Statistics 3.1, Statistics 3.2

A-level. This model illustrates selected content; references are not a claim of full coverage or exam-board endorsement.

Useful starting knowledge: Fractions; Two-way tables; Intersection of events.

AQA specification · Edexcel specification