When a question gives the result of a draw and asks which bag was chosen, keep every path that could have produced the result. Divide the probability of the requested path by the total probability of observing that result.
The 2027 A-Level H2 Mathematics syllabus (9758) includes simple conditional probabilities and using tree diagrams to calculate probabilities. This original example shows why the chance of choosing each bag before the draw matters after the result is known.
- Write the chance of selecting each bag before writing the colour probabilities.
- Multiply along each path to the observed result.
- Add the relevant paths to form the denominator.
Worked example | A red counter appears
A machine selects Bag A with probability 1/4 and Bag B with probability 3/4. Bag A has four red counters and one blue counter. Bag B has two red counters and three blue counters. One counter is drawn uniformly from the selected bag. Given that it is red, find the probability that Bag A was selected.
On the tree, the A-to-red path has probability (1/4) × (4/5) = 1/5. The B-to-red path has probability (3/4) × (2/5) = 3/10. These are disjoint routes, so the total chance of drawing red is 1/5 + 3/10 = 1/2.
| Path | Bag selection | Red given bag | Joint probability |
|---|---|---|---|
| A → red | 1/4 | 4/5 | 1/5 |
| B → red | 3/4 | 2/5 | 3/10 |
The required conditional probability is (1/5) ÷ (1/2) = 2/5. Bag A has the larger red share, but the machine chooses Bag B more often. In this example a red result is still more likely to have come from Bag B.
Why adding the red fractions gives the wrong denominator
The red fractions 4/5 and 2/5 describe draws from different bags. Before adding, weight each by the chance that its bag was selected. Adding the unweighted fractions would give 6/5, which cannot be the probability of observing red.
A frequency model gives the same result. Imagine 100 trials divided according to the probabilities. About 25 select A and 75 select B. The expected red counts are 20 from A and 30 from B. Among the 50 expected red results, the share from A is 20/50. These are expected counts for illustration.
Change the observation or the starting chances
If the observed counter is blue instead, the joint probabilities are (1/4) × (1/5) = 1/20 from A and (3/4) × (3/5) = 9/20 from B. The total probability of blue is 1/2 and the conditional probability of A given blue is 1/10.
Now return to a red draw and make the bag choices equally likely. The weighted red paths become 2/5 and 1/5, so the conditional probability of A given red rises to 2/3. Changing the starting chances changes the inference even when the bags contain the same counters.
A study partner can point at the final colour and ask which paths remain. Read the numerator and denominator aloud before using the calculator.
Frequently asked questions
Why is the answer different from 4/5?
The fraction 4/5 describes a red draw from Bag A. The question asks about the bag after red has been observed.
Can a table replace the tree?
Yes. Make sure the entries represent joint probabilities or counts from a consistent population.
Official information
SEAB’s 2027 A-Level H2 Mathematics syllabus (9758), Section B, Probability. Checked 8 November 2026.