Probability is one of the friendliest topics in O-Level E Math Paper 2. The questions are usually worth 6 to 10 marks, the working is short, and the method almost never changes. The catch is the phrase "without replacement", which trips up more students than any other three words in the topic. Here is exactly how to handle it.
What "without replacement" actually changes
When an item is drawn and not put back, two things happen on the second draw:
- The total number of items goes down by one.
- The count of the item you just drew goes down by one.
That is it. The denominators shrink, and one numerator shrinks. Students who lose marks here almost always remember to change the denominator but forget the numerator, or the other way round.
Compare the two setups with 5 red and 3 blue marbles in a bag:
- With replacement: second draw is always out of 8.
- Without replacement: second draw is always out of 7, and the branch you came from has one fewer.
A full worked example
A bag contains 5 red marbles and 3 blue marbles. Two marbles are drawn at random, one after the other, without replacement.
Step 1: Draw the tree and label it properly.
First draw branches: R with $\frac{5}{8}$, B with $\frac{3}{8}$.
Second draw, following R: R with $\frac{4}{7}$, B with $\frac{3}{7}$.
Second draw, following B: R with $\frac{5}{7}$, B with $\frac{2}{7}$.
Notice every second-stage denominator is 7. If you see a 7 and an 8 mixed together in the second stage, something has gone wrong.
Step 2: Multiply along branches, add between branches.
$P(\text{both red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$
$P(\text{both blue}) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56} = \frac{3}{28}$
$P(\text{one of each colour}) = \frac{5}{8} \times \frac{3}{7} + \frac{3}{8} \times \frac{5}{7} = \frac{15}{56} + \frac{15}{56} = \frac{15}{28}$
Step 3: Check that the four end-branches sum to 1.
$\frac{20}{56} + \frac{15}{56} + \frac{15}{56} + \frac{6}{56} = \frac{56}{56} = 1$
This ten-second check catches nearly every arithmetic slip. Do it every time.
The three question types that keep appearing
1. "At least one" questions
$P(\text{at least one blue}) = 1 - P(\text{no blue}) = 1 - \frac{5}{14} = \frac{9}{14}$
Using the complement is faster and safer than adding three branches. Write the line "$= 1 - P(\text{both red})$" so the examiner sees the method even if your arithmetic slips.
2. "The same colour" or "different colours" questions
Same colour means add the both-red and both-blue branches. Different colours is the complement of that, so $1 - \left(\frac{5}{14} + \frac{3}{28}\right) = \frac{15}{28}$. If you compute both ways and they agree, you are safe.
3. "The second one is red" questions
This one surprises people. You must add across both first-draw branches:
$P(\text{2nd is red}) = \frac{5}{8} \times \frac{4}{7} + \frac{3}{8} \times \frac{5}{7} = \frac{35}{56} = \frac{5}{8}$
The answer equals the probability the first is red. That is a genuine result, not a coincidence, and a good sanity check.
Common mistakes that cost marks
- Adding when you should multiply. Along a path means multiply. Between paths means add.
- Reducing the wrong numerator. After drawing blue, it is the blue count that drops, not the red.
- Converting to decimals too early. Keep fractions. $\frac{5}{8} \times \frac{4}{7}$ is exact, while 0.63 times 0.57 invites rounding penalties.
- Unlabelled trees. If the question says "Draw a tree diagram", marks are awarded for the labels and the probabilities on every branch, not the shape.
- Three-stage questions. For three draws, the third-stage denominators are 6. Only draw the branches you actually need for the required outcome.
How to present it for full marks
Examiners want to see the multiplication before the answer. Write:
$$P(\text{both red}) = \frac{5}{8} \times \frac{4}{7} = \frac{5}{14}$$
Not just $\frac{5}{14}$. The method mark is often worth as much as the answer mark, and it survives a slip in simplifying. Give final answers as fractions in lowest terms unless the question asks for decimals or a specific number of significant figures.
A quick drill for the week before the exam
Take any past-paper probability question with replacement and rewrite it without replacement. Solve both versions side by side. After five or six pairs, the adjustment to the denominators becomes automatic, which is exactly what you want under time pressure in Paper 2.
If tree diagrams still feel shaky, or you want someone to check your working line by line, come and try a lesson with our tutors who sat these papers recently and know the mark schemes inside out. Start a free trial class with Dojo on WhatsApp.