Laws of Probability and Combined Events
Combined events questions ask for the probability of two or more things happening together, and two laws of probability do almost all of the work: add the probabilities for mutually exclusive events, multiply them for independent events. On this page you will learn when each law applies, how the second pick changes when nothing is replaced, and how the complement shortcut unlocks "at least one" problems in Edexcel IGCSE Maths. Then lock it in with the auto-marked practice questions below: five rooms of randomly generated problems, from single-law starters to multi-stage exam-style challenges.
Choosing the Right Law
Almost every combined events question comes down to one decision: are you being asked about this event or that event, or this event and that event? The word tells you which law to use.
- Read for the key word. "Either A or B happens" points to the addition law. "Both A and B happen" points to the multiplication law.
- Check the condition. The OR law needs the events to be mutually exclusive (they cannot happen at the same time). The AND law needs the events to be independent (one has no effect on the other).
- Add or multiply the probabilities. OR means add; AND means multiply.
- Sanity check. Every probability must land between 0 and 1. If your answer does not, a law was applied where its condition failed.
Mutually Exclusive or Not?
Two events are mutually exclusive when they cannot happen together: one counter cannot be red and green at the same time. The OR law only works when the two circles below have no overlap.
Mutually exclusive: no overlap
Not mutually exclusive: an overlap exists
On the left, adding \(P(A) + P(B)\) is safe. On the right, simply adding would count the highlighted overlap twice, so the OR law cannot be used directly.
The Laws of Probability for Combined Events
Only for mutually exclusive events: events that cannot happen at the same time.
Only for independent events: one event has no effect on the other.
\(A'\) means "not A". An event either happens or it does not, so the two probabilities add to 1.
When a list of outcomes covers every possibility, their probabilities add to exactly 1. This finds a missing value in a table.
\(P(\text{at least one}) = 1 - P(\text{none})\)
Working out "none" takes one multiplication; listing every way to get "at least one" takes many. Always go through the complement.
Without Replacement: the Second Pick Changes
When an item is taken out and not put back, the second pick is a conditional probability: the bag has changed. Both numbers in the second fraction react to the first pick: the total drops by one, and the count of the colour that left drops by one.
A bag holds 5 red and 3 blue counters, and two counters are taken out one after the other without replacement. The tree diagram shows every branch.
After a red counter leaves, only 4 of the remaining 7 are red. After a blue counter leaves, 5 of the remaining 7 are red: every second-pick fraction has denominator 7.
To find the probability that both counters are red, multiply along the top branch:
\[ \begin{array}{rcl} P(R, R) &=& \dfrac{5}{8} \times \dfrac{4}{7} \\ &=& \dfrac{20}{56} \\ &=& \dfrac{5}{14} \end{array} \]To find the probability of one counter of each colour, add the two routes that give it (the routes are mutually exclusive, so the OR law applies):
\[ \begin{array}{rcl} P(\text{one of each}) &=& \dfrac{5}{8} \times \dfrac{3}{7} + \dfrac{3}{8} \times \dfrac{5}{7} \\ &=& \dfrac{15}{56} + \dfrac{15}{56} \\ &=& \dfrac{30}{56} \\ &=& \dfrac{15}{28} \end{array} \]Worked Examples
💡 Example 1: The OR Law and a Missing Probability
A four-sided spinner is biased. The table shows the probability of each score. Find the missing probability, then find the probability of spinning a 1 or a 4.
| Score | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Probability | \(\dfrac{3}{10}\) | \(\dfrac{1}{5}\) | ? | \(\dfrac{1}{10}\) |
The four scores cover every outcome, so the probabilities add to 1:
\[ \begin{array}{rcl} P(3) &=& 1 - \left( \dfrac{3}{10} + \dfrac{1}{5} + \dfrac{1}{10} \right) \\ &=& 1 - \dfrac{6}{10} \\ &=& \dfrac{2}{5} \end{array} \]The scores are mutually exclusive, so:
\[ \begin{array}{rcl} P(1 \text{ or } 4) &=& \dfrac{3}{10} + \dfrac{1}{10} \\ &=& \dfrac{4}{10} \\ &=& \dfrac{2}{5} \end{array} \]What's happening?
A spinner must land on exactly one score, so the whole table adds to 1. Whatever the three known fractions leave behind is the missing probability.
One spin cannot score 1 and 4 at the same time, so the OR law simply adds the two probabilities.
💡 Example 2: The AND Law with a Complement
The probability that Zoe's bus arrives on time on any morning is \(\dfrac{7}{10}\), independently of other mornings. For two mornings, find the probability that the bus is (a) on time both mornings, (b) on time on the first morning but late on the second.
Multiply along a branch. Both mornings show the same fractions, which is what \(\text{independent}\) means.
(a) The mornings are independent:
\[ \begin{array}{rcl} P(T, T) &=& \dfrac{7}{10} \times \dfrac{7}{10} \\ &=& \dfrac{49}{100} \end{array} \](b) First find \(P(\text{late})\) with the complement: \(1 - \dfrac{7}{10} = \dfrac{3}{10}\). Then:
\[ \begin{array}{rcl} P(T, L) &=& \dfrac{7}{10} \times \dfrac{3}{10} \\ &=& \dfrac{21}{100} \end{array} \]What's happening?
"Both" means the first AND the second, so multiply.
In (b) the second event is "late", which is the complement of "on time". Convert it first, then multiply as usual.
💡 Example 3: At Least One, Without Replacement
A pack of 9 game cards contains 6 star cards and 3 moon cards. Two cards are taken at random without replacement. Find the probability of getting at least one star card.
The highlighted route is moon then moon, the only one with no star. Example 3 subtracts it from 1.
Go through the complement: the only way to get no star is moon then moon.
\[ \begin{array}{rcl} P(\text{no star}) &=& \dfrac{3}{9} \times \dfrac{2}{8} \\ &=& \dfrac{6}{72} \\ &=& \dfrac{1}{12} \end{array} \] \[ \begin{array}{rcl} P(\text{at least one star}) &=& 1 - \dfrac{1}{12} \\ &=& \dfrac{11}{12} \end{array} \]What's happening?
"At least one star" covers three different routes, but "no star" is a single route. Work out the easy one and subtract from 1.
The second fraction is conditional: after one moon card leaves, 2 moons remain out of 8 cards.
💡 Example 4: A Multi-Stage Problem (Three Trials)
Each time Ravi plays an arcade game, the probability that he wins a token is \(\dfrac{1}{4}\), independently of other plays. He plays three times. Find the probability that he wins (a) exactly two tokens, (b) at least one token.
Each row is one order. The three products are identical, which is where the \(\times 3\) comes from.
(a) Exactly two wins happens by three routes: WWL, WLW and LWW. Each route has the same product, so multiply one route by 3:
\[ \begin{array}{rcl} P(\text{exactly two}) &=& 3 \times \dfrac{1}{4} \times \dfrac{1}{4} \times \dfrac{3}{4} \\ &=& 3 \times \dfrac{3}{64} \\ &=& \dfrac{9}{64} \end{array} \](b) Use the complement: "at least one win" is everything except "no wins".
\[ \begin{array}{rcl} P(\text{at least one}) &=& 1 - \dfrac{3}{4} \times \dfrac{3}{4} \times \dfrac{3}{4} \\ &=& 1 - \dfrac{27}{64} \\ &=& \dfrac{37}{64} \end{array} \]What's happening?
This question uses both laws at once: multiply along each route (AND), then add the routes together (OR), because different routes cannot both happen.
A full three-stage tree has eight branches, but you only multiply along the routes the question asks about: three routes in (a) and one in (b).
Losing a play has probability \(1 - \dfrac{1}{4} = \dfrac{3}{4}\), the complement of winning.
🔑 Key Points
- "Or" means add, but only when the events are mutually exclusive.
- "And" means multiply, but only when the events are independent.
- \(P(A') = 1 - P(A)\): the complement turns "at least one" into one quick calculation.
- A full table of outcomes always adds to 1; that finds any missing probability.
- Without replacement, both parts of the second fraction change: total down by 1, and the colour that left down by 1.
- Multiply along a tree branch; add between whole branches.
⚠️ Common Pitfalls
- Adding probabilities for events that can happen together: the overlap gets counted twice.
- Multiplying when the events are not independent, for example two picks from the same bag without replacement.
- Forgetting to change the denominator (and the numerator) on the second pick when nothing is replaced.
- Treating "at least one" as "exactly one": they are different events with different answers.
- An answer below 0 or above 1 is impossible: it always signals a wrong law or wrong arithmetic.