Tree Diagrams for Dependent Events

Draw a tree diagram when the first item is not put back, so the second branches change, then multiply along paths, add across paths and solve for a missing number.

In the exam for Cambridge IGCSE Maths 0580, Edexcel IGCSE Maths 4MA1, Cambridge O Level Maths 4024 · 81 real past-paper questions below

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When the Second Branch Changes

Some questions say the first item is not put back. Then the second pick comes from a smaller bag, so its probabilities change.

The 2-mark question: complete the tree. A bag has 5 black and 3 white counters. Mia takes one, keeps it, then takes another.

Worked example

1
$first: \frac{5}{8}, \frac{3}{8}$ — There are 8 counters to start: 5 black and 3 white. Each probability is how many of that colour over how many in the bag.
2
$after black: \frac{4}{7}, \frac{3}{7}$ — One black is gone, so 4 black are left. The bag now holds 7, so 7 goes on the bottom. All 3 white are still there.
3
$after white: \frac{5}{7}, \frac{2}{7}$ — One white is gone, so 2 white are left out of 7. All 5 black are still there.
4
$\frac{4}{7} + \frac{3}{7} = 1, \frac{5}{7} + \frac{2}{7} = 1$ — Check each pair adds to 1. The second counter must be black or white, so nothing is missing.

Rule

Not put back: the bottom number drops by 1, and the colour you took drops by 1. Each pair of branches still adds to 1.

Common mistake

Copying the first branches onto the second set. That is only right when the item is put back, as in Tree Diagrams for Independent Events.

Two Picks: Multiply, Then Add

The 3-mark question: two items taken, find both, one of each, or the same colour. Multiply along each path, using the changed second branch. Then add the paths that fit.

Worked example

1
$B, W = \frac{5}{8} \times \frac{3}{7}$ — Path one: black first (5 of 8), then white. After a black, all 3 white are still there out of 7. Along a path you multiply.
2
$= \frac{15}{56}$ — Tops times tops: 5 × 3 = 15. Bottoms times bottoms: 8 × 7 = 56.
3
$W, B = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56}$ — Path two: white first, then black. All 5 black are still there, out of 7.
4
$\frac{15}{56} + \frac{15}{56} = \frac{30}{56}$ — One of each can happen either way round, so add the two paths.
5
$= \frac{15}{28}$ — Divide top and bottom by 2 to simplify.

Same colour works the same way: both black plus both white, $\frac{20}{56} + \frac{6}{56} = \frac{13}{28}$ .

Common mistake

Writing $\frac{5}{8} \times \frac{5}{8}$ for both black. The first black is gone, so the second is $\frac{4}{7}$ .

Key idea

Write every path as a product, like $\frac{5}{8} \times \frac{3}{7}$ , before you add. Most of the marks are for those products, so they still score if you slip later.

Three Colours: Different Colours

The most common 4-mark question: three colours in the bag, and you want two different colours. Six paths give different colours but only three give the same, so find the same and take it from 1.

Formula

P (different) = 1 - P (same)

“Same” means both red, both green or both blue. Every pair is either the same or different, so the two probabilities add to 1.

A bag has 4 red, 3 green and 2 blue counters. Two are taken, not put back. Find the probability they are different colours.

Worked example

1
$RR = \frac{4}{9} \times \frac{3}{8} = \frac{12}{72}$ — Both red: 4 of the 9 are red, then 3 red are left out of 8. Multiply along the path.
2
$GG = \frac{3}{9} \times \frac{2}{8} = \frac{6}{72}$ — Both green: 3 of 9, then 2 green left out of 8.
3
$BB = \frac{2}{9} \times \frac{1}{8} = \frac{2}{72}$ — Both blue: 2 of 9, then only 1 blue left out of 8.
4
$P (same) = \frac{12 + 6 + 2}{72} = \frac{20}{72}$ — Add the three same-colour paths. Each one is a different way to get a match.
5
$P (different) = 1 - \frac{20}{72} = \frac{52}{72}$ — Every other path has two different colours, so take the same-colour total from 1: 72 − 20 = 52.
6
$= \frac{13}{18}$ — Divide top and bottom by 4 to simplify.

Common mistake

Adding only red-green, red-blue and green-blue. Red then green and green then red are different paths, so each pair of colours counts twice.

Key idea

The long way also works: add all six different-colour paths. It gives $\frac{52}{72}$ too, but it is easy to miss one.

Numbers on Cards: Sums and Products

Some questions put numbers on the cards and ask for a total or a product. List the pairs that work, then treat each pair as a path.

Five cards show 1, 2, 3, 3 and 4. Two are taken, not put back. Find the probability that the two numbers add to 6.

Worked example

1
$2 + 4, 4 + 2, 3 + 3$ — List every pair that makes 6, in order. On a tree, 2 then 4 and 4 then 2 are two paths. The 1 cannot work: there is no 5.
2
$P (2, 4) = \frac{1}{5} \times \frac{1}{4} = \frac{1}{20}$ — One card shows 2, out of 5. The 4 is still there, out of the 4 cards left.
3
$P (4, 2) = \frac{1}{5} \times \frac{1}{4} = \frac{1}{20}$ — The same cards in the other order give the same answer.
4
$P (3, 3) = \frac{2}{5} \times \frac{1}{4} = \frac{2}{20}$ — Two cards show 3. After one 3 is taken, only one 3 is left out of 4.
5
$\frac{1}{20} + \frac{1}{20} + \frac{2}{20} = \frac{4}{20}$ — Any of the three paths gives 6, so add them.
6
$= \frac{1}{5}$ — Divide top and bottom by 4 to simplify.

Common mistake

Using $\frac{2}{5} \times \frac{2}{5}$ for 3 then 3. The first 3 is not put back, so only one 3 is left out of 4.

Key idea

Products work the same way. For a positive product, list the pairs with two positive numbers or two negative numbers, because a negative times a negative is positive.

Three Picks and At Least One

The 3 or 4 mark question: three items taken, not put back. Each pick has a smaller bag. For at least one, take the no path from 1.

Formula

P (at least one) = 1 - P (none)

A bag has 3 black and 7 white counters. Three are taken. Find the probability that at least one is black.

Worked example

1
$P (no black) = \frac{7}{10} \times \frac{6}{9} \times \frac{5}{8}$ — No black means white, white, white. Each time a white goes, the white count and the bag both drop by 1.
2
$= \frac{210}{720}$ — Tops: 7 × 6 × 5 = 210. Bottoms: 10 × 9 × 8 = 720.
3
$= \frac{7}{24}$ — Divide top and bottom by 30. This is the only path with no black.
4
$1 - \frac{7}{24} = \frac{17}{24}$ — Every other path has at least one black, so take the no-black path from 1.

Common mistake

Cubing $\frac{7}{10}$ . Each pick comes from a smaller bag, so the fractions are all different.

Key idea

Exactly one black: the black can be first, second or third. Each path is $\frac{126}{720}$ , so the answer is $3 \times \frac{126}{720} = \frac{21}{40}$ .

Key idea

“All the black counters are still in the bag” means none of the picks was black. That is P(none) itself, so do not take it from 1.

Find the Missing Number

The 5 or 6 mark question: the bag has $n$ items and you are given one probability. Write the path with letters, then solve the quadratic.

A bag has $n$ counters. 3 are black, the rest are white. Two are taken. The probability both are white is $\frac{7}{15}$ . Find $n$ .

Worked example

1
$P (1st white) = \frac{n - 3}{n}$ — 3 of the n counters are black, so n − 3 are white. The first pick is from all n.
2
$P (2nd white) = \frac{n - 4}{n - 1}$ — One white is gone, so n − 4 white are left, out of n − 1.
3
$\frac{n - 3}{n} \times \frac{n - 4}{n - 1} = \frac{7}{15}$ — Both white is one path, so multiply and set it equal to the probability you are given.
4
$15 (n - 3) (n - 4) = 7 n (n - 1)$ — Cross multiply: times both sides by 15 and by n(n − 1) to clear the fractions.
5
$15 n^{2} - 105 n + 180 = 7 n^{2} - 7 n$ — Expand. (n − 3)(n − 4) = n² − 7n + 12, then times 15. On the right, 7n times n − 1.
6
$8 n^{2} - 98 n + 180 = 0$ — Take 7n² from both sides and add 7n to both sides, so the right side is 0.
7
$4 n^{2} - 49 n + 90 = 0$ — Divide every term by 2 to make the numbers smaller.
8
$(n - 10) (4 n - 9) = 0$ — Factorise. Check: −10 × −9 = 90 and −9n − 40n = −49n.
9
$n = 10 or n = \frac{9}{4}$ — Set each bracket equal to 0.
10
$n = 10$ — Reject $\frac{9}{4}$ : you cannot have a quarter of a counter, so n must be a whole number.

Common mistake

Using $n$ on both picks. After one is taken there are $n - 1$ left, so the second bottom is $n - 1$ .

Key idea

Check: $\frac{7}{10} \times \frac{6}{9} = \frac{42}{90} = \frac{7}{15}$ . It works.

Which Method? Read the Question

Look for the words that say if the item goes back, and how many paths you need.

You see	You do
Put back, or two bags	Tree Diagrams for Independent Events
Not put back, or takes two	second branches change
Both, or two of a kind	multiply along one path
One of each, same colour	multiply each path, then add
Three colours, different colours	$1 - P (same)$
Numbers on the cards	list the pairs, each pair is a path
At least one	$1 - P (none)$
Bag has n items	write the path, solve the quadratic
Two chosen from a table or histogram	find the counts, then it is a bag
If this happens, then that…	Conditional Probability

Key idea

"Takes two at random" means not put back, even if the question never says it.

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