Combined events
A combined event is two or more things happening one after another, such as flipping a coin twice or drawing two counters from a bag. A tree diagram shows the first event as a set of branches, then grows a fresh set of branches from the end of each one for the second event.
- Tree diagram
- A branching diagram where each branch carries the probability of one step. Every full path from left to right is one possible combined outcome.
The picture below is written in text form. Flipping a fair coin twice gives four paths:
Each branch here has probability \(\tfrac{1}{2}\), and the two branches leaving any single point always add to \(1\). That is a quick self-check: if a pair of branches does not total \(1\), one of them is wrong.
Multiplying along branches
To find the probability of a complete path, multiply the probabilities along its branches from left to right. This works because you want the first step and the second step to happen.
For two flips of a fair coin, the chance of heads then heads is:
All four paths give \(\tfrac{1}{4}\), and \(\tfrac{1}{4} + \tfrac{1}{4} + \tfrac{1}{4} + \tfrac{1}{4} = 1\), so nothing has been missed.
Along means times, across means plus
Multiply along a single path to get that path's probability. Add across different paths when the event can happen in more than one way. Mixing these two up is the most common slip in the whole topic.
With and without replacement
The key question before building the second set of branches: is the first item put back? This decides whether the second-step probabilities change.
| With replacement | Without replacement | |
|---|---|---|
| Total each draw | Stays the same | Drops by one |
| Second-step probability | Unchanged (independent) | Changes (dependent) |
| Bag of 5 red, 3 blue | \(P(\text{red}) = \tfrac{5}{8}\) both draws | \(\tfrac{5}{8}\) then \(\tfrac{4}{7}\) |
A bag holds 5 red and 3 blue counters. Two are drawn without replacement. Find the probability both are red.
Adding across paths
Some events are satisfied by more than one path. To handle these, find each path by multiplying, then add the paths together. "One of each colour", for instance, is satisfied by red then blue or blue then red.
Continuing the bag of \(5\) red and \(3\) blue without replacement:
As a full check, the three cases cover everything: \(P(RR) = \tfrac{20}{56}\), \(P(\text{one of each}) = \tfrac{30}{56}\), \(P(BB) = \tfrac{3}{8}\times\tfrac{2}{7} = \tfrac{6}{56}\). Together \(\tfrac{20 + 30 + 6}{56} = \tfrac{56}{56} = 1\).
Where this is assessed
Setting up correct branch probabilities and combining them is Criterion A (Knowing and understanding). A fully labelled tree with a probability on every branch is Criterion C (Communicating). Using it to judge a real question, such as whether a game is fair, reaches Criterion D (Applying in real-life contexts).
Check yourself
Try each before opening the answer.
1. A fair coin is flipped twice. Find the probability of exactly one head. +
Exactly one head happens on two paths: heads then tails, or tails then heads. Each is \(\tfrac{1}{2} \times \tfrac{1}{2} = \tfrac{1}{4}\). Add them: \(\tfrac{1}{4} + \tfrac{1}{4} = \) \(\dfrac{1}{2}\).
2. From the bag of 5 red and 3 blue, two are drawn without replacement. Find the probability both are blue. +
First blue: \(\dfrac{3}{8}\). Then \(2\) blue remain of \(7\): \(\dfrac{2}{7}\). Multiply along the path: \(\dfrac{3}{8} \times \dfrac{2}{7} = \dfrac{6}{56} = \) \(\dfrac{3}{28}\).
3. A spinner lands on green with probability 0.2 each spin. It is spun twice, with each spin independent. Find the probability of green on both spins. +
The spins are independent, so the probability stays \(0.2\) each time. Multiply along the path: \(0.2 \times 0.2 = \) \(0.04\).
Part of the Standard Maths library. Spotted an error or want a topic added? That feedback makes the notes better.