Combined & Conditional ProbabilityEdexcel IGCSE Maths: Revision notes

Two events are independent if the outcome of one has no effect on the outcome of the other. For independent events A and B, use the AND rule:

$P(A \text{ and } B) = P(A) \times P(B)$

Example: rolling a die and flipping a coin. $P(6) = \frac{1}{6}$, $P(\text{heads}) = \frac{1}{2}$, so $P(6 \text{ and heads}) = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12}$.

For mutually exclusive events (they cannot both happen), use the OR rule:

$P(A \text{ or } B) = P(A) + P(B)$

Example: picking a card that is a King or a Queen: $P(K) + P(Q) = \frac{4}{52} + \frac{4}{52} = \frac{8}{52} = \frac{2}{13}$.

A tree diagram shows all possible outcomes of two or more events in sequence. Each branch is labelled with a probability, and probabilities on branches from the same point always sum to 1.

To find the probability of a specific outcome (a path through the tree), multiply along the branches. To find the probability of several different outcomes, add the results of each relevant path.

Worked example (numerically, no diagram): A bag has 3 red and 2 blue counters. One counter is drawn, replaced, then a second is drawn.

• P(red then red) = $\frac{3}{5} \times \frac{3}{5} = \frac{9}{25}$
• P(red then blue) = $\frac{3}{5} \times \frac{2}{5} = \frac{6}{25}$
• P(blue then red) = $\frac{2}{5} \times \frac{3}{5} = \frac{6}{25}$
• P(blue then blue) = $\frac{2}{5} \times \frac{2}{5} = \frac{4}{25}$

All four outcomes sum to $\frac{25}{25} = 1$ — always check this as a sense-check.

P(one of each colour, in any order) = P(red then blue) + P(blue then red) = $\frac{6}{25} + \frac{6}{25} = \frac{12}{25}$.

When an item is not replaced, the total number of items decreases and the probabilities on the second draw change depending on what was drawn first. This makes the second event dependent on the first.

Worked example: A bag has 3 red and 2 blue counters. One is drawn and NOT replaced, then a second is drawn.

• P(red then red) = $\frac{3}{5} \times \frac{2}{4} = \frac{6}{20} = \frac{3}{10}$ (only 2 red left out of 4 total)
• P(red then blue) = $\frac{3}{5} \times \frac{2}{4} = \frac{6}{20} = \frac{3}{10}$
• P(blue then red) = $\frac{2}{5} \times \frac{3}{4} = \frac{6}{20} = \frac{3}{10}$
• P(blue then blue) = $\frac{2}{5} \times \frac{1}{4} = \frac{2}{20} = \frac{1}{10}$

Check: $\frac{3}{10} + \frac{3}{10} + \frac{3}{10} + \frac{1}{10} = \frac{10}{10} = 1$.

P(both same colour) = P(RR) + P(BB) = $\frac{3}{10} + \frac{1}{10} = \frac{4}{10} = \frac{2}{5}$.

Conditional probability is the probability of an event occurring given that another event has already happened. It is written $P(A \mid B)$, read as "the probability of A given B".

The formula (given in the formula sheet, but understanding it matters more than memorising it):

$P(A \mid B) = \frac{P(A \text{ and } B)}{P(B)}$

Most IGCSE questions do not require the formula directly — they ask you to reason through a without-replacement or restricted scenario using a tree diagram or a simple count.

Worked example: A bag has 5 red and 3 blue counters. A counter is drawn and not replaced. Given that the first counter drawn was red, what is the probability the second is also red?

After removing one red, there are 4 red and 3 blue left, 7 total. So $P(\text{red}_2 \mid \text{red}_1) = \frac{4}{7}$.

Worked example with a table: 100 students are asked if they study French. 60 study French, of whom 24 are boys. There are 45 boys in total. Given a student is a boy, P(studies French) = $\frac{24}{45} = \frac{8}{15}$ (only considering the 45 boys as the new total).

Two-way tables organise data into rows and columns (e.g. gender vs subject choice) and are a common way IGCSE questions test conditional probability without mentioning trees.

To find $P(A \mid B)$ from a table: look only at the row or column for B, then find what fraction of that row/column also satisfies A. You are restricting the total (denominator) to the given condition, not the whole table.

Example table (counts):

P(plays sport | girl) = $\frac{18}{30} = \frac{3}{5}$ — use the girls' total (30), not 60.

P(girl | plays sport) = $\frac{18}{43}$ — use the "plays sport" total (43), not 60.