Probability and statistics · GCSE Maths
Tree diagrams
GCSE Maths tree diagrams: multiply along a branch for AND, add complete pathways for OR, and change the second fraction without replacement, with full working shown.
AND multiplies along a branch. OR adds complete pathways. Without replacement, the second fraction changes — show the new denominator on the tree.
The important bits
What you need to know
- 1
Each fork of a tree must add to 1. If the first-draw branches are 3/5 and 2/5, those two fractions already use up the whole probability at that stage.
- 2
Multiply along a branch for both events happening (AND). P(red then red) = (3/5) × (2/4) without replacement, not 3/5 + 2/4.
- 3
Add complete pathways for OR when the pathways cannot happen together: P(one of each colour) = P(red then blue) + P(blue then red).
- 4
Without replacement, the bag has changed. After a red is taken from 3 red and 2 blue, 4 counters remain, so the next red is 2/4, not 3/5.
- 5
With replacement, or independent events, the second-stage fractions copy the first. Two coin tosses: each fork is 1/2, and P(two heads) = 1/2 × 1/2 = 1/4.
- 6
“At least one” is often 1 minus the pathway of none. P(at least one red in two draws) = 1 − P(blue then blue). That is usually fewer products than listing every mixed pathway.
- 7
Keep fractions until the last line unless the question asks for a decimal. Cancel before multiplying: 3/8 × 2/7 = 6/56 = 3/28.
- 8
A finished tree is not the answer. The question asks for a probability; the tree is the working. Box the products you add, and check that all mutually exclusive complete outcomes sum to 1.
Quotations worth analysing
Short evidence. Real method.
“Multiply along the branches; add the outcomes”
This pair of verbs is the whole topic. Multiplying gives AND on one path. Adding combines mutually exclusive paths for OR.
“Without replacement”
The denominator falls by 1, and the numerator of the colour just drawn also falls by 1. Copying the first-stage fractions is the standard lost mark.
“P(A and B) = P(A) × P(B given A)”
On a tree, “B given A” is the second-stage fraction on the A branch. Do not multiply by P(A) a second time if the question already says “given that the first was red”.
Go deeper
Build the tree, then multiply, then add
A bag contains 3 red and 2 blue counters. Two counters are drawn without replacement. First fork: 3/5 red, 2/5 blue. Second fork after red: 2/4 red, 3/4 blue. After blue: 3/4 red, 1/4 blue. Each pair from a point adds to 1, which is a check you can do before any products. P(both red) = 3/5 × 2/4 = 6/20 = 3/10. P(one of each) = 3/5 × 3/4 + 2/5 × 3/4. Wait: after red, blue is 2/4 not 3/4 — 2 blue still there, 4 left, so 2/4. After blue, red is 3/4. So 3/5 × 2/4 + 2/5 × 3/4 = 6/20 + 6/20 = 12/20 = 3/5. Adding 3/5 and 2/4 would mix a first-draw probability with a second-draw probability from different worlds. Only add products of complete pathways.
Go deeper
At least one, and “given that”
P(at least one red) = 1 − P(blue then blue) = 1 − (2/5 × 1/4) = 1 − 2/20 = 18/20 = 9/10. Listing both red, red then blue, and blue then red should give the same 9/10. Conditional language changes where you start. “Given that the first is red” means you are already on the red first-branch; P(second is red | first is red) is just 2/4, not (3/5)×(2/4). “Given that at least one is red” is harder: restrict to the pathways that have a red, then among those the probability of both red is P(both red) / P(at least one red) = (3/10) / (9/10) = 1/3. Higher papers ask this. The tree still helps because the products are the numerators and denominators of that fraction.
Go deeper
Three stages still multiply along and add across
Three coins, or three draws, do not change the verbs. P(three heads) = (1/2)³ = 1/8. P(exactly two heads) = three pathways HHT, HTH, THH, each 1/8, total 3/8. Without replacement from a larger bag, write every second- and third-stage fraction with the new totals. If a question forbids a tree, the same products still appear in a sentence: “3/8 × 2/7 × 1/6”. Expected frequency is probability × number of trials: if P(both red) = 3/28 and this experiment is repeated 56 times, expect 6 double-reds. That is an average, not a promise. End with a size check: every probability you report must sit between 0 and 1. A 1.2 means you added along a branch instead of multiplying.
See the idea in action
A bag contains 3 red and 5 blue counters. Two counters are drawn without replacement. Find P(both red) and P(one of each colour). Step 1: First draw: P(red) = 3/8, P(blue) = 5/8. Step 2: Without replacement, after red: P(red) = 2/7, P(blue) = 5/7. After blue: P(red) = 3/7, P(blue) = 4/7. Step 3: P(both red) = 3/8 × 2/7 = 6/56 = 3/28. (AND: multiply along the red–red branch.) Step 4: P(one of each) = 3/8 × 5/7 + 5/8 × 3/7 = 15/56 + 15/56 = 30/56 = 15/28. (OR: add the two mixed pathways.) Check: P(both blue) = 5/8 × 4/7 = 20/56 = 5/14 = 10/28, and 3/28 + 15/28 + 10/28 = 1.
Exam technique
Turn knowledge into marks
Label every branch with a fraction, not “likely”. Multiply along, add across. If the bag is without replacement, the second denominator must be one smaller than the first.
Common mistakes
Do not give these marks away
- 01
Using the same second-draw fraction after taking an item out, as if the bag had been replaced.
- 02
Adding along a branch instead of multiplying, which often produces a probability greater than 1.
- 03
Adding the first-stage fraction to a second-stage fraction from a different pathway, instead of adding complete products.
A bag has 3 red and 2 blue counters. Two are drawn without replacement. P(both red) is
A3/5 × 2/4
B3/5 × 3/5
C3/5 + 2/4
D3/5 × 2/5
Show the answer
3/5 × 2/4. After one red is taken, 2 red remain out of 4. 3/5 × 3/5 is with replacement. 3/5 + 2/4 adds along the branch. 3/5 × 2/5 keeps the old denominator of 5.
Quick questions
If this is the bit you searched
When do I multiply probabilities on a tree?
When you want both events on one pathway (AND). Change the second fraction if the first event is without replacement or otherwise dependent.
When do I add probabilities on a tree?
When you want either of two (or more) complete pathways (OR), and those pathways cannot happen together. Add the products, not the little fractions on different forks.
How does without replacement change the tree?
The total on the second fork is one less, and the colour just drawn is one less. Write the new fraction on that branch; do not copy the first stage.
How do I find P(at least one)?
Often 1 minus the pathway with none. P(at least one red) = 1 − P(no reds). Check it matches the sum of all pathways that contain a red.