Combined Probability and Tree Diagrams
The Skill
Combined Events
When two or more events happen, we can find probabilities using:
- AND (both events) → multiply probabilities
- OR (either event) → add probabilities (if mutually exclusive)
Tree Diagrams
Tree diagrams show all possible outcomes for a sequence of events.
How to Draw a Tree Diagram
- Draw branches for the first event
- From each branch, draw branches for the second event
- Write probabilities on each branch
- Probabilities from any point must sum to 1
Reading a Tree Diagram
- Multiply along branches to find the probability of that path
- Add probabilities of paths that give the outcome you want
Independent Events
Events are independent if one doesn't affect the other. $$P(A \text{ and } B) = P(A) \times P(B)$$
Dependent Events (Without Replacement)
If the first event changes the probabilities for the second event, adjust the fractions.
Example: A bag has 3 red and 2 blue balls. Two are drawn without replacement.
- First ball: P(red) = $\frac{3}{5}$
- If first was red: P(second red) = $\frac{2}{4}$ (one less red, one less total)
The AND/OR Rules
$$P(A \text{ and } B) = P(A) \times P(B)$$ $$P(A \text{ or } B) = P(A) + P(B)$$ (for mutually exclusive events)
The Traps
Common misconceptions and how to avoid them.
Adding probabilities when you should multiply "The AND/OR Mixup"
The Mistake in Action
A coin is flipped twice. Find the probability of getting two heads.
Wrong: $P(HH) = \frac{1}{2} + \frac{1}{2} = 1$
Why It Happens
Students confuse when to add and when to multiply. They may think "two events" means add, without considering whether it's AND or OR.
The Fix
AND = Multiply (both events must happen) OR = Add (either event can happen)
"Two heads" means Head AND Head: $$P(HH) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$$
Memory aid:
- "AND" makes it harder (probability gets smaller) → multiply
- "OR" gives more options (probability gets larger) → add
Check: Getting two heads should be less likely than getting one head!
Spot the Mistake
Find P(two heads) for two coin flips
$P(HH) = \frac{1}{2} + \frac{1}{2} = 1$
Click on the line that contains the error.
Not changing fractions for without replacement "The Replacement Forget"
The Mistake in Action
A bag contains 4 red and 6 blue counters. Two counters are taken without replacement. Find the probability of getting two red.
Wrong: $P(RR) = \frac{4}{10} \times \frac{4}{10} = \frac{16}{100}$
Why It Happens
Students use the same fractions for both events, forgetting that "without replacement" means the total number changes.
The Fix
Without replacement means the first pick affects the second.
After taking one red counter:
- Red counters left: $4 - 1 = 3$
- Total counters left: $10 - 1 = 9$
$$P(RR) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$$
With replacement would use $\frac{4}{10}$ twice. Without replacement — update the numbers!
Spot the Mistake
4 red, 6 blue. Two taken WITHOUT replacement. Find P(RR).
$P(RR) = \frac{4}{10} \times \frac{4}{10}$
Click on the line that contains the error.
Tree diagram branches not summing to 1 "The Branch Blunder"
The Mistake in Action
A spinner has probability 0.6 of landing on blue. Draw a tree diagram for spinning twice.
Wrong: First spin: P(Blue) = 0.6, P(Not Blue) = 0.6
Why It Happens
Students may copy the same probability for all branches or calculate the complement incorrectly.
The Fix
From any point, the branches must sum to 1.
If P(Blue) = 0.6, then P(Not Blue) = $1 - 0.6 = 0.4$
Check each "fork" in your tree:
- Does P(Blue) + P(Not Blue) = 1?
- $0.6 + 0.4 = 1$ ✓
This applies at every level of the tree!
Spot the Mistake
P(Blue) = 0.6 for each spin
P(Blue) = 0.6, P(Not Blue) = 0.6
Click on the line that contains the error.
The Deep Dive
Apply your knowledge with these exam-style problems.
Level 1: Fully Worked
Complete solutions with commentary on each step.
Question
A fair coin is flipped twice. Draw a tree diagram and find the probability of getting exactly one head.
Solution
Tree Diagram:
First flip Second flip Outcome Probability
H -------- H HH 1/2 × 1/2 = 1/4
/
/ ---- T HT 1/2 × 1/2 = 1/4
/
Start
\ ---- H TH 1/2 × 1/2 = 1/4
\
T -------- T TT 1/2 × 1/2 = 1/4
Exactly one head: HT or TH
$$P(\text{exactly one head}) = P(HT) + P(TH)$$ $$= \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$$
Answer: $\frac{1}{2}$
Question
A spinner has P(Red) = 0.3 and P(Blue) = 0.7. It is spun twice. Find the probability of getting the same colour both times.
Solution
Outcomes with same colour: Red-Red OR Blue-Blue
$$P(RR) = 0.3 \times 0.3 = 0.09$$
$$P(BB) = 0.7 \times 0.7 = 0.49$$
$$P(\text{same colour}) = P(RR) + P(BB)$$ $$= 0.09 + 0.49 = 0.58$$
Answer: 0.58
Check: P(different) should be $1 - 0.58 = 0.42$ P(RB) + P(BR) = $0.3 \times 0.7 + 0.7 \times 0.3 = 0.21 + 0.21 = 0.42$ ✓
Question
The probability that it rains on any day is 0.2. Find the probability that it rains on at least one of the next two days.
Solution
Method 1: Add all success paths At least one rain day: RR, RN, NR (where R = rain, N = no rain)
$P(RR) = 0.2 \times 0.2 = 0.04$ $P(RN) = 0.2 \times 0.8 = 0.16$ $P(NR) = 0.8 \times 0.2 = 0.16$
$P(\text{at least one}) = 0.04 + 0.16 + 0.16 = 0.36$
Method 2: Use complement (easier!) P(at least one) = 1 − P(none)
$P(NN) = 0.8 \times 0.8 = 0.64$ $P(\text{at least one}) = 1 - 0.64 = 0.36$
Answer: 0.36
Level 2: Scaffolded
Fill in the key steps.
Question
A bag contains 5 red and 3 green counters. Two counters are drawn without replacement. Find the probability of getting two red counters.
Level 3: Solo
Try it yourself!
Question
A biased dice has P(six) = 0.2. The dice is rolled three times. Find the probability of getting exactly two sixes.
Show Solution
P(six) = 0.2, P(not six) = 0.8
Outcomes with exactly two sixes:
- SSN: $0.2 \times 0.2 \times 0.8 = 0.032$
- SNS: $0.2 \times 0.8 \times 0.2 = 0.032$
- NSS: $0.8 \times 0.2 \times 0.2 = 0.032$
$$P(\text{exactly two sixes}) = 0.032 + 0.032 + 0.032 = 0.096$$
Answer: 0.096
(Note: There are 3 ways to arrange 2 sixes and 1 non-six)
Examiner's View
Mark allocation: Simple combined events: 2-3 marks. Tree diagrams: 3-5 marks.
Common errors examiners see:
- Adding when you should multiply (and vice versa)
- Not adjusting probabilities for "without replacement"
- Branches not summing to 1
- Missing outcomes
What gains marks:
- Label your tree diagram clearly
- Show probabilities on branches
- Show multiplication for each path
- List all paths that match the outcome
AQA Notes
AQA often uses contexts like spinners, counters, or two-stage experiments.