For the probability of exactly one red counter given at least one red counter, divide the probability of exactly one red by the probability of at least one red. The numerator can simplify this way because every outcome with exactly one red already satisfies the condition.
Conditional probability and counting methods are in the 2027 A-Level H2 Mathematics 9758 syllabus. This example uses two draws without replacement and compares three questions that sound similar but give different denominators.
- Write down the event after ‘given’ and use it to set the denominator.
- Keep outcomes with one or two reds when the condition says at least one.
- Count both draw orders when exactly one red can appear in either position.
- Use the same representation throughout, with weighted tree paths or equally likely individual pairs.
CHAPTER 1 OF 4 · Back to contents
Translate the condition before counting
Let A be the event that exactly one counter is red and B the event that at least one counter is red. The formula is P(A | B) = P(A ∩ B) ÷ P(B), provided P(B) > 0. Since A is contained in B, A ∩ B = A.
With two draws of red or blue counters, the condition B retains the colour paths RR, RB and BR. The target A retains RB and BR. Counting two of the three colour paths as 2/3 assumes that the paths have equal probabilities. Compute their weights or count equally likely pairs of individual counters.
The statement ‘at least one is red’ leaves the position unknown. ‘The first counter is red’ identifies a position and therefore selects a different set of outcomes.
CHAPTER 2 OF 4 · Back to contents
Worked example | Use both paths and their probabilities
A bag contains three red and five blue counters. Two are drawn uniformly at random without replacement. Given that at least one is red, find the probability that exactly one is red.
| Path | Probability | Retained by the condition? |
|---|---|---|
| RR | 3/8 × 2/7 = 6/56 | Yes |
| RB | 3/8 × 5/7 = 15/56 | Yes |
| BR | 5/8 × 3/7 = 15/56 | Yes |
| BB | 5/8 × 4/7 = 20/56 | No |
The probability of exactly one red is 30/56. The probability of at least one red is 36/56. The conditional probability is (30/56) ÷ (36/56) = 5/6. You can also find the denominator by subtracting P(BB) from 1.
For a combinations check, there are 15 individual red-blue pairs and 3 red-red pairs. All individual unordered pairs are equally likely because each can be drawn in two orders with equal probability. Restrict the count to the 18 pairs containing a red, then take 15/18 = 5/6.
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CHAPTER 3 OF 4 · Back to contents
Change the wording and solve again
If the first counter is known to be red, exactly one red requires the second counter to be blue. Of the seven remaining counters, five are blue, so this conditional probability is 5/7. Write the condition beside the answer to keep it distinct from 5/6.
If you are given exactly one red and asked for the probability that the first counter is red, the denominator is P(RB) + P(BR). The numerator is P(RB). Both paths have probability 15/56, so the answer is 1/2.
For a fresh check, take four red and four blue counters and use the original wording. There are 16 red-blue pairs and 6 red-red pairs, so the probability is 16/(16 + 6) = 8/11. Compare the new counts with the original 15 and 3.
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CHAPTER 4 OF 4 · Back to contents
Frequently asked questions
Why does the denominator include RR?
Two reds satisfy the information that at least one counter is red. RR stays in the denominator even though it fails the target of exactly one.
When can I cancel the total number of pairs?
You can cancel it when both event probabilities use the same equally likely sample space. With an unordered count, use unordered individual pairs in both event counts.
How can I catch a swapped condition?
Read the vertical bar as ‘given’. Put the event on the right beneath the fraction bar, and check that your numerator describes outcomes that also meet that event.
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