Conditional Probability
Conditional probability asks how a probability changes when you are given extra information. The key is to identify the group that is still possible, then calculate the required proportion within that group.
Higher. This lesson brings together Sample Spaces and Two-Way Tables, Venn Diagrams and Set Notation, and Tree Diagrams and Combined Events. Their basic representations and without-replacement calculations are already taught in the shared Foundation & Higher lessons.
“Given” changes the group you count from
P(A given B) means the probability of A among outcomes where B has happened. You may see the shorthand P(A | B), read “the probability of A given B”. The event after the bar determines the denominator.
Conditional probability = number in both A and B ÷ number in B. Restrict the sample space to B first. This requires B to have positive probability; a group containing no possible outcomes cannot be used as a denominator.
| Year | Cycle | Walk | All |
|---|---|---|---|
| 10 | 9 | 11 | 20 |
| 11 | 6 | 14 | 20 |
| All | 15 | 25 | 40 |
Worked example — condition in two different directions
Choose a student uniformly from these 40. Given that the student cycles, there are only 15 possible students; 9 are in Year 10. P(Year 10 given cycles) = 9/15 = 3/5.
Given that the student is in Year 10, the relevant total is 20. P(cycles given Year 10) = 9/20. The two directions need not have the same probability.
Compare the unconditional joint probability: P(Year 10 and cycles) = 9/40. “Given” and “and” ask different questions.
Condition inside a Venn diagram
Among 30 students, art only has 11, both art and music has 7, music only has 8, and neither has 4. Let A mean art and B mean music.
Worked example — restrict to a circle or its complement
Given music, the possible students are the 7 + 8 = 15 in B. P(art given music) = 7/15.
Given not music, the possible students are the 11 + 4 = 15 outside B. P(art given not music) = 11/15.
Given at least one subject, the possible group has 11 + 7 + 8 = 26 students. P(both given at least one) = 7/26. The condition decides the denominator each time.
Expected frequencies make reverse questions visible
Supplier A provides 60% of a shop’s items and supplier B provides 40%. An item from A has a 10% chance of damage; an item from B has a 25% chance. Choose an item at random from this mix.
In the table, D means damaged and N means not damaged. Build a convenient model of 200 items. Expect 120 from A and 80 from B. Of A’s items, 0.1 × 120 = 12 are damaged; of B’s, 0.25 × 80 = 20 are damaged.
| Source | D | N | All |
|---|---|---|---|
| A | 12 | 108 | 120 |
| B | 20 | 60 | 80 |
| All | 32 | 168 | 200 |
Worked example — reverse the direction of the given information
Given damage, only the 32 damaged items remain possible. Of those, 12 came from A: P(A given damage) = 12/32 = 3/8 = 0.375.
This is not the 0.1 probability of damage given A. Reversing the words reverses which group forms the denominator.
Given no damage, P(B) = 60/168 = 5/14. These are model probabilities derived from expected counts, not a claim that an actual batch must contain exactly those counts.
Read the same conditional problem from a tree
In this tree D means damaged and N means not damaged. The second-stage probabilities describe damage status given the supplier.
Worked example — use joint paths before restricting
P(A and D) = 0.6 × 0.1 = 0.06. P(B and D) = 0.4 × 0.25 = 0.10. Therefore P(D) = 0.06 + 0.10 = 0.16.
Among the damaged outcomes, the fraction from A is 0.06/0.16 = 3/8, matching the expected-frequency table.
In general, P(A given B) = P(A and B) ÷ P(B), provided P(B) > 0. Rearranging gives P(A and B) = P(A given B) × P(B). A probability tree uses exactly this idea when you multiply along a path.
Without replacement: use the information actually given
A bag has 3 red and 2 blue counters; two are drawn without replacement. If you know the first is red, the chance the second is red is 2/4. That is a forward branch. A different question tells you the second is red and asks about the first.
Worked example — condition on a later result
P(RR) = 3/5 × 2/4 = 3/10. P(BR) = 2/5 × 3/4 = 3/10. These are the only paths ending in red, so P(second red) = 6/10.
P(first red given second red) = (3/10)/(6/10) = 1/2. Restricting to the relevant complete paths prevents you from assuming that an unrelated branch probability answers the question.
Test whether events are independent
For P(B) > 0, A and B are independent exactly when P(A given B) = P(A). Equivalently, their joint probability equals P(A) × P(B). Knowing B then leaves A’s chance unchanged.
Worked example — compare within-group and whole-group probabilities
In a group of 100, 40 belong to A, 30 belong to B and 12 belong to both. P(A given B) = 12/30 = 0.4, equal to P(A) = 40/100 = 0.4. Therefore A and B are independent in this model.
In the art/music example, P(art given music) = 7/15 differs from P(art) = 18/30 = 3/5. Those events are dependent.
Mutually exclusive events with positive probabilities are not independent: if B occurs, A becomes impossible. In real sampled data, a difference in observed proportions is evidence to consider; it does not by itself prove a universal causal relationship.
Think Like a Mathematician
Why can 10% of A’s items be damaged while 37.5% of damaged items came from A? The same damaged-from-A outcomes sit inside two different groups. Drawing the whole table makes both denominators visible.
Common Mistakes
Using the grand total after a condition: first restrict to the group you have been told about.
Reversing the condition: P(A given B) need not equal P(B given A).
Dividing a branch by the wrong total: a reverse tree question often needs several complete paths in its denominator.
Confusing dependence with cause: a changed conditional probability does not by itself establish causation.
Practice
1. Among 60 pupils, 24 cycle to school and 9 of those cyclists are in a club. Find P(club given cycles).
The condition restricts the group to 24 cyclists. Probability = 9/24 = 3/8.
2. There are 15 club members among those 60 pupils. Find P(cycles given club) and P(cycles and club).
Given club: 9/15 = 3/5. AND among all pupils: 9/60 = 3/20.
3. A Venn diagram has A only 10, both 6, B only 9 and neither 5. Find P(A given B′).
Outside B are A only and neither: 10 + 5 = 15. Of these, 10 are in A, so 10/15 = 2/3.
4. Supplier X provides 40% of items with damage probability 0.05. Supplier Y provides the rest with damage probability 0.10. Find P(X given damaged).
For 500 model items, expect 200 from X and 300 from Y: damaged counts 10 and 30. Among 40 damaged items, 10 are from X, so P = 1/4. Equivalently, (0.4×0.05)/(0.4×0.05+0.6×0.10) = 0.02/0.08.
5. A bag has 4 red and 2 blue counters. Draw two without replacement. Find P(first blue given second red).
BR = 2/6 × 4/5 = 4/15. RR = 4/6 × 3/5 = 2/5 = 6/15. The second-red total is 10/15. Thus (4/15)/(10/15) = 2/5.
6. P(A)=0.4, P(B)=0.5 and P(A and B)=0.2. Are A and B independent?
P(A given B) = 0.2/0.5 = 0.4, equal to P(A), so yes. Equivalently, 0.4 × 0.5 = 0.2.
7. P(A given B)=0.7 and P(B)=0.3. Find P(A and B).
0.7 × 0.3 = 0.21. The condition does not justify treating 0.7 as the unconditional probability of A.
8. A group has no members in B. Can you calculate P(A given B) by dividing the count in A and B by the count in B?
No. The denominator would be zero. There are no possible outcomes in the given group, so this conditional probability is undefined.
Frequently Asked Questions
Is every dependent tree a Higher conditional-probability question?
No. Following dependent branches, such as updating counts without replacement, is shared content. Higher conditional problems ask you to restrict or reinterpret the possible outcomes using given information, often across tables, Venn diagrams or complete tree paths.
Do I need to memorise a separate reverse-probability rule?
No. Find the outcomes satisfying the condition, then the proportion of those that also satisfy the requested event. Counts, expected frequencies or complete path probabilities all give the same reasoning.