Probability in IGCSE Maths
From the probability scale to tree diagrams and conditional probability — with clear diagrams, worked examples and an exam-style question. Learn the AND and OR rules, handle 'with' and 'without replacement', and set out your working to earn full marks.
0580 · 8.4The probability scale
Every probability is a number between \(0\) and \(1\). An impossible event has probability \(0\); a certain event has probability \(1\). For equally likely outcomes:
\[P(\text{event}) = \frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}\]The probabilities of all the possible outcomes always add up to \(1\).
The probability that an event does not happen is:
\[P(\text{not }A) = 1 - P(A)\]This is one of the most useful shortcuts in probability — especially for 'at least one' questions.
A fair six-sided die is rolled. Find the probability of (a) rolling a \(5\), and (b) not rolling a \(5\).
(a) One of the six outcomes is a \(5\):
\[P(5) = \frac{1}{6}\](b) Use the complement:
\[P(\text{not }5) = 1 - \frac{1}{6} = \frac{5}{6}\]Experimental probability
When outcomes are not equally likely — a biased spinner, a drawing pin landing point up — we estimate probability from data using relative frequency:
\[P(\text{event}) \approx \frac{\text{number of times the event happened}}{\text{total number of trials}}\]The more trials you carry out, the closer the relative frequency gets to the true probability.
To predict how often an event will happen, multiply its probability by the number of trials:
\[\text{Expected frequency} = P(\text{event}) \times \text{number of trials}\]A biased spinner lands on red with probability \(0.3\). It is spun \(200\) times. How many times would you expect it to land on red?
Combining events: AND and OR
Two rules cover almost every probability question:
AND → multiply. For independent events (one does not affect the other):
\[P(A \text{ and } B) = P(A) \times P(B)\]OR → add. For mutually exclusive events (they cannot both happen):
\[P(A \text{ or } B) = P(A) + P(B)\]On a tree diagram you multiply along the branches of a single path (AND), and add between the separate paths that satisfy the question (OR). Keep this phrase in your head and most tree questions fall into place.
A fair coin is flipped and a fair die is rolled. Find the probability of getting a head and a six.
These are independent, so multiply:
\[P(\text{head and }6) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}\]Tree diagrams: independent events
A tree diagram shows every outcome of a two-stage experiment. Each branch is labelled with its probability, and the branches leaving any single point always add up to \(1\). Here a bag holds \(3\) red and \(2\) blue counters. A counter is taken, its colour noted, and it is replaced — so the second pick is independent of the first.
Multiply along each path to get the probability of that outcome. For example:
\[P(\text{red, red}) = \frac{3}{5} \times \frac{3}{5} = \frac{9}{25}\]Using the tree above, find the probability of getting one red and one blue, in any order.
Two paths give one of each — (red, blue) and (blue, red) — so add their probabilities:
\[P(\text{one of each}) = \frac{6}{25} + \frac{6}{25} = \frac{12}{25}\]Tree diagrams: without replacement
ExtendedIf the first counter is not replaced, the second pick depends on the first: there is one fewer counter, and one fewer of whichever colour was taken. The second set of branches changes to match.
With \(3\) red and \(2\) blue, after taking a red there are \(2\) red and \(2\) blue left (\(4\) in total), so the next branch is \(\frac{2}{4}\). Multiply along the path as before:
\[P(\text{red, red}) = \frac{3}{5} \times \frac{2}{4} = \frac{6}{20} = \frac{3}{10}\]Every time you take a counter without replacement, the total drops by one. If your second-stage denominators are still the same as the first stage, you have forgotten to remove the counter.
Conditional probability
ExtendedConditional probability is the probability of an event given that something has already happened. A 'without replacement' tree is conditional probability in action: each second-stage branch is the probability for the second pick given the first. You rarely need formal notation at IGCSE — just adjust the numerator and denominator to fit the new situation.
In a group of \(40\) students, \(15\) of the \(22\) girls study French and \(8\) of the \(18\) boys study French. A student who studies French is chosen at random. Find the probability that the student is a boy.
The total number studying French is \(15 + 8 = 23\). Of these, \(8\) are boys, so restrict to that group:
\[P(\text{boy} \mid \text{French}) = \frac{8}{23}\]Exam-style question
A bag contains \(5\) red sweets and \(3\) yellow sweets. Two sweets are taken at random, without replacement.
(a) Find the probability that both sweets are red. [2]
(b) Find the probability that the two sweets are different colours. [3]
(c) Find the probability that at least one sweet is yellow. [2]
Show solution (a)
Multiply along the red–red path:
\[P(\text{red, red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}\]Show solution (b)
Two paths give different colours — (red, yellow) and (yellow, red) — so add them:
\[P(\text{different}) = \frac{5}{8}\times\frac{3}{7} + \frac{3}{8}\times\frac{5}{7} = \frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}\]Show solution (c)
Use the complement — 'at least one yellow' is the opposite of 'no yellow' (both red):
\[P(\text{at least one yellow}) = 1 - \frac{20}{56} = \frac{36}{56} = \frac{9}{14}\]This is a very common Paper 2 and Paper 4 structure. Part (c) is the giveaway: whenever you see 'at least one', reach for \(1 - P(\text{none})\) rather than adding several cases — it is faster and avoids mistakes. Note too that the denominators drop from \(8\) to \(7\) because the first sweet is not replaced.
Common mistakes
Along a single path (this and then that), you multiply. Adding is only for combining separate paths (this or that). Mixing the two up is the most common tree-diagram error.
Without replacement, the second-stage denominator must drop by one. Leaving it unchanged treats the events as independent when they are not.
The probabilities on the branches leaving any single point must add up to \(1\). If they do not, one of them is wrong — a quick check that catches many slips.
For 'at least one', use \(1 - P(\text{none})\). Adding up every case that contains 'at least one' is slow and easy to get wrong.
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