IGCSE Probability Tree Diagram Without Replacement: Worked Example

A fairground game uses a box containing 10 keyrings: 6 are silver and 4 are gold.

A player takes a keyring from the box at random, keeps it (it is not replaced), and then takes a second keyring from the box at random.

The situation can be modelled by a probability tree diagram with this structure:

First pick has two branches:
- Silver with probability $\tfrac{6}{10}$
- Gold with probability $\tfrac{4}{10}$
Second pick has two branches growing from each first‑pick branch, leading to the outcomes Silver–Silver, Silver–Gold, Gold–Silver, Gold–Gold. The probabilities on these second‑pick branches must be worked out, because one keyring has already been removed.

(a) Draw the tree diagram and label the probabilities on the four second‑pick branches. [2]

(b) Find the probability that both keyrings are silver. [2]

(d) Find the probability that the player takes at least one gold keyring. [1]

Setting up the tree

There are $10$ keyrings: $6$ silver (S) and $4$ gold (G). The keyring is not replaced, so for the second pick only $9$ keyrings remain, and the make‑up depends on what was taken first.

(a) Second‑pick branch probabilities

If the first keyring was silver (S): $5$ silver and $4$ gold remain out of $9$ .

$P(\text{S}\mid\text{S}) = \frac{5}{9}, \qquad P(\text{G}\mid\text{S}) = \frac{4}{9}$

If the first keyring was gold (G): $6$ silver and $3$ gold remain out of $9$ .

$P(\text{S}\mid\text{G}) = \frac{6}{9} = \frac{2}{3}, \qquad P(\text{G}\mid\text{G}) = \frac{3}{9} = \frac{1}{3}$

Check: each pair of second‑pick branches sums to $1$ (e.g. $\tfrac{5}{9}+\tfrac{4}{9}=1$ ). ✓

The complete tree therefore carries these probabilities:

First pick	Second pick	Path probability
S $\left(\tfrac{6}{10}\right)$	S $\left(\tfrac{5}{9}\right)$	$\tfrac{6}{10}\times\tfrac{5}{9}=\tfrac{30}{90}=\tfrac13$
S $\left(\tfrac{6}{10}\right)$	G $\left(\tfrac{4}{9}\right)$	$\tfrac{6}{10}\times\tfrac{4}{9}=\tfrac{24}{90}=\tfrac{4}{15}$
G $\left(\tfrac{4}{10}\right)$	S $\left(\tfrac{6}{9}\right)$	$\tfrac{4}{10}\times\tfrac{6}{9}=\tfrac{24}{90}=\tfrac{4}{15}$
G $\left(\tfrac{4}{10}\right)$	G $\left(\tfrac{3}{9}\right)$	$\tfrac{4}{10}\times\tfrac{3}{9}=\tfrac{12}{90}=\tfrac{2}{15}$

(The four path probabilities sum to $\tfrac13+\tfrac{4}{15}+\tfrac{4}{15}+\tfrac{2}{15}=\tfrac{5}{15}+\tfrac{4}{15}+\tfrac{4}{15}+\tfrac{2}{15}=\tfrac{15}{15}=1$ . ✓)

(b) Both silver

Multiply along the Silver → Silver path:

$P(\text{S, S}) = \frac{6}{10}\times\frac{5}{9} = \frac{30}{90} = \boxed{\dfrac{1}{3}}$

(c) Exactly one gold

“Exactly one gold” means one gold and one silver, in either order, so add the two mixed paths:

$P(\text{exactly one gold}) = P(\text{S, G}) + P(\text{G, S})$

$= \underbrace{\frac{6}{10}\times\frac{4}{9}}_{=\,\frac{24}{90}} \;+\; \underbrace{\frac{4}{10}\times\frac{6}{9}}_{=\,\frac{24}{90}} = \frac{24}{90}+\frac{24}{90} = \frac{48}{90} = \boxed{\dfrac{8}{15}}$

(d) At least one gold

The quickest method is the complement: “at least one gold” is the opposite of “no gold at all”, i.e. the opposite of both silver found in part (b).

$P(\text{at least one gold}) = 1 - P(\text{both silver}) = 1 - \frac{1}{3} = \boxed{\dfrac{2}{3}}$

Alternative (adding paths): $P(\text{S,G})+P(\text{G,S})+P(\text{G,G}) = \tfrac{4}{15}+\tfrac{4}{15}+\tfrac{2}{15} = \tfrac{10}{15} = \tfrac{2}{3}$ , the same answer, confirming the result.

Final answers

(a) $\tfrac{5}{9},\ \tfrac{4}{9}$ (after silver) and $\tfrac{2}{3},\ \tfrac{1}{3}$ (after gold)
(b) $\dfrac{1}{3}$
(c) $\dfrac{8}{15}$
(d) $\dfrac{2}{3}$