Interpreting Conditional Probability and Expected Frequencies Revision Notes | GCSE AQA Higher Maths

1. Why Conditional Probability Changes Everything

Here is a question that trips up a lot of students: a bag contains 3 red and 2 blue counters. You pick one and keep it. What is the probability the second counter is red? The answer depends entirely on what the first counter was — and that dependence is exactly what conditional probability captures.

Conditional probability is not a new formula to memorise. It is a shift in perspective: instead of asking 'what is the probability of this event?', you ask 'what is the probability of this event, now that I know something else?' That extra information shrinks the sample space, and the probability changes accordingly.

At GCSE, you will not need to use the formal conditional probability formula. Instead, the examiner expects you to read conditional probabilities from three representations: two-way tables, tree diagrams, and Venn diagrams. Each one makes the restricted sample space visible in a different way.

2. Two-Way Tables and Conditional Probability

A two-way table is the most direct tool for conditional probability. The trick is knowing which total to use as your denominator. When a condition is given, your denominator is the row total or column total for that condition — not the grand total.

The table above shows results for 50 students. To find $P(\text{Passed} \mid \text{Studied})$, you restrict attention to the 'Studied' row only. There are $28 + 7 = 35$ students who studied, and 28 of them passed.

Worked example: Reading Conditional Probability from a Two-Way Table

Using the table above, calculate $P(\text{Passed} \mid \text{Studied})$ and $P(\text{Studied} \mid \text{Passed})$.

1. Condition: 'Studied' — restrict to the Studied row. — The condition tells you which row or column becomes your new sample space. [M1]
2. $28 + 7 = 35$ students studied in total. — This is the denominator for any probability conditional on 'Studied'.
3. 28 students studied AND passed. — This is the numerator — the count satisfying both the condition and the event.
4. $P(\text{Passed} \mid \text{Studied}) = \dfrac{28}{35} = \dfrac{4}{5}$ — Simplify the fraction. [A1]
5. Condition is now 'Passed' — restrict to the Passed column. Column total: $28 + 6 = 34$. Favourable: 28 studied and passed. $P(\text{Studied} \mid \text{Passed}) = \dfrac{28}{34} = \dfrac{14}{17}$ — Notice the denominator changed — the condition changed, so the restricted sample space changed. [A1]

Answer: $P(\text{Passed} \mid \text{Studied}) = \dfrac{4}{5}$; $P(\text{Studied} \mid \text{Passed}) = \dfrac{14}{17}$

Notice that $P(\text{Passed} \mid \text{Studied}) eq P(\text{Studied} \mid \text{Passed})$. Swapping the condition and the event gives a completely different probability. This is one of the most common errors on GCSE Higher exam papers.

3. Tree Diagrams for Dependent Events

When events happen in sequence and the outcome of the first affects the second, a tree diagram makes the conditional probabilities explicit. Each second-stage branch already has the condition built in — the probabilities on those branches are conditional probabilities.

The key rule for tree diagrams: multiply along branches to find the probability of a combined outcome, and add the results for different routes to the same overall outcome.

In the tree above, the bag starts with 3 red and 2 blue counters (5 total). After a red is removed, only 4 counters remain — 2 red and 2 blue. After a blue is removed, 4 remain — 3 red and 1 blue. The second-branch probabilities change depending on what happened first: that is conditional probability in action.

Worked example: Using a Tree Diagram to Find a Conditional Probability

A bag has 3 red and 2 blue counters. Two counters are picked without replacement. Given that the second counter is red, find the probability that the first counter was also red.

1. Route 1: Red then Red (RR) Route 2: Blue then Red (BR) — These are the only two routes that end with a red second counter. The condition restricts us to these routes only. [M1]
2. $P(RR) = \dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{6}{20}$ — First pick: 3 red out of 5. Second pick: 2 red remain out of 4 total remaining.
3. $P(BR) = \dfrac{2}{5} \times \dfrac{3}{4} = \dfrac{6}{20}$ — First pick: 2 blue out of 5. Second pick: 3 red remain out of 4 total remaining.
4. $P(\text{2nd is red}) = \dfrac{6}{20} + \dfrac{6}{20} = \dfrac{12}{20}$ — Add the two routes — this is the denominator for the conditional probability. [M1]
5. $P(\text{1st red} \mid \text{2nd red}) = \dfrac{P(RR)}{P(\text{2nd red})} = \dfrac{6/20}{12/20} = \dfrac{6}{12} = \dfrac{1}{2}$ — Divide the favourable route by the total probability of the condition. [A1]

Answer: $P(\text{1st red} \mid \text{2nd red}) = \dfrac{1}{2}$

4. Venn Diagrams and Conditional Probability

A Venn diagram shows how two sets overlap. For conditional probability, the logic is the same as with a two-way table: the condition tells you which region becomes your new total, and you count how much of the event falls inside that region.

Worked example: Finding Conditional Probability from a Venn Diagram

Using the Venn diagram above (40 students total), calculate $P(T \mid F)$ and interpret your answer.

1. Condition: plays football ($F$). Restrict to the entire football circle. — The football circle contains all students for whom the condition is true. [M1]
2. $15 + 8 = 23$ students play football. — This is the denominator — the size of the restricted sample space.
3. 8 students play both football and tennis (the intersection). — These are the students who satisfy both the condition (plays football) and the event (plays tennis).
4. $P(T \mid F) = \dfrac{8}{23}$ — Numerator: intersection. Denominator: total in the condition's circle. [A1]
5. $\dfrac{8}{23} \approx 0.348$ — Among students who play football, approximately 34.8% also play tennis. Always relate the answer back to the context when asked to interpret. [A1]

Answer: $P(T \mid F) = \dfrac{8}{23}$. Among students who play football, the probability they also play tennis is $\dfrac{8}{23}$.

5. Expected Frequencies and Interpreting Conditional Probability

Probabilities can feel abstract. Expected frequencies make them concrete by asking: if this situation happened $n$ times, how many times would we expect this outcome? You calculate expected frequency as probability multiplied by the number of trials.

Expected frequencies are particularly useful when interpreting conditional probability in context. If $P(T \mid F) = \dfrac{8}{23}$ and there are 230 football players in a school, you would expect $\dfrac{8}{23} \times 230 = 80$ of them to also play tennis.

Worked example: Using Expected Frequencies to Interpret Conditional Probability

A survey of 200 people finds that 120 drink coffee. Of those, 72 also drink tea. A person is chosen at random from all 200. Given they drink coffee, find the probability they also drink tea. Then use expected frequencies to interpret this for a group of 500 coffee drinkers.

1. Condition: drinks coffee. Total coffee drinkers = 120. Coffee AND tea drinkers = 72. — Restrict the sample space to coffee drinkers only. [M1]
2. $P(\text{tea} \mid \text{coffee}) = \dfrac{72}{120} = \dfrac{3}{5}$ — Simplify: $72 \div 24 = 3$ and $120 \div 24 = 5$. [A1]
3. $\text{Expected frequency} = \dfrac{3}{5} \times 500$ — Multiply the conditional probability by the new group size of 500 coffee drinkers.
4. $\dfrac{3}{5} \times 500 = \dfrac{3 \times 500}{5} = \dfrac{1500}{5} = 300$ — Among 500 coffee drinkers, we would expect 300 to also drink tea. [A1]

Answer: $P(\text{tea} \mid \text{coffee}) = \dfrac{3}{5}$. In a group of 500 coffee drinkers, 300 would be expected to also drink tea.

6. Choosing the Right Representation

GCSE Higher exam questions will sometimes give you the data and ask you to choose how to represent it. Here is a practical guide: use a two-way table when you have two categorical variables and frequency data already organised in rows and columns. Use a tree diagram when events happen in a clear sequence, especially without replacement. Use a Venn diagram when the question focuses on overlapping sets or membership.

• Two-way table: two categorical variables, data given as counts — read off conditional probabilities by fixing a row or column.
• Tree diagram: sequential events, especially without replacement — second-branch probabilities are already conditional.
• Venn diagram: overlapping sets — restrict to one circle to apply the condition.

7. Exam Technique and Common Mistakes

Conditional probability questions on GCSE Higher exam papers tend to follow a predictable pattern: you are given a representation (table, tree, or Venn diagram), asked to calculate a probability, and then asked to interpret it. The marks split roughly as one mark for the correct method (right denominator) and one mark for the correct simplified answer.

When a question says 'given that', underline those words. They are telling you the condition — and the condition determines your denominator. Using the grand total instead of the conditional total is the single most penalised error in this topic.

All three representations — two-way tables, tree diagrams, and Venn diagrams — are tools for making the restricted sample space visible. Once you can see which group the condition locks you into, the conditional probability is a straightforward fraction: favourable outcomes within that group, divided by the total within that group.