POP

E8.3 Probability of Combined Events: Sample Spaces, Venn & Tree Diagrams

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Combined events involve two or more things happening together, such as rolling two dice or picking two counters from a bag.

01 — Explanation

Which diagram should I use for a given combined-event question?

Diagram	Best used for
Sample space (grid)	Two independent events shown together, e.g. two dice, listing every outcome
Venn diagram	Events described with set language (intersection, union, complement)
Tree diagram	A sequence of events, especially where probabilities change between stages (without replacement)

02 — Explanation

How do I use a sample space diagram?

List every possible outcome of the first event along one axis of a grid and every possible outcome of the second event along the other axis, then fill each cell with the combined result (a sum, product, or pair). Once complete, count the cells that satisfy the event you want and divide by the total number of cells.

Two fair dice are rolled and their scores added. There are $36$ equally likely cells in the grid, and $6$ of them give a total of $7$ ( $1+6,2+5,3+4,4+3,5+2,6+1$ ), so $P(\text{total}=7)=\dfrac{6}{36}=\dfrac{1}{6}$ .

03 — Explanation

How do I fill in a tree diagram for events without replacement?

Each time an item is removed and not put back, both the number of favourable outcomes and the total number remaining decrease by $1$ — so every branch after the first must be recalculated based on what happened before it.

A bag contains $5$ green and $4$ yellow counters. Two are drawn without replacement.

First draw: $P(G_1)=\dfrac{5}{9}$ , $\quad P(Y_1)=\dfrac{4}{9}$
Second draw, given green first (now $8$ left, $4$ green): $P(G_2\mid G_1)=\dfrac{4}{8}$
Second draw, given yellow first (now $8$ left, $5$ green): $P(G_2\mid Y_1)=\dfrac{5}{8}$

Exam Tip

Writing $P(G_2\mid G_1)=\dfrac{4}{9}$ instead of $\dfrac{4}{8}$ is the single most common error with "without replacement" trees — the denominator must shrink too, not just the numerator.

04 — Explanation

How is a "with replacement" tree different?

If each item is put back before the next draw, the total and the composition of the bag never change, so every second-draw branch uses exactly the same probabilities as the first draw. For the bag above but with replacement, $P(G_2\mid G_1)=P(G_2\mid Y_1)=\dfrac{5}{9}$ throughout — there's no need to track what happened on the first draw at all.

05 — Explanation

How do I combine probabilities along and across the branches of a tree?

Multiply along a single path (this represents one event AND the next). Add the results of separate paths that both lead to the outcome you actually want (this represents one path OR another).

Finding $P(\text{both same colour})$ for the without-replacement bag above:

P(G,G) = \frac{5}{9}\times\frac{4}{8} = \frac{20}{72} \qquad\qquad P(Y,Y) = \frac{4}{9}\times\frac{3}{8} = \frac{12}{72}

P(\text{same colour}) = \frac{20}{72}+\frac{12}{72} = \frac{32}{72} = \frac{4}{9}

Worked Example

A box contains $6$ red pens and $3$ blue pens. Two pens are chosen at random without replacement. Find the probability that at least one pen is blue.

It's usually faster to find $P(\text{no blue})$ first and use the complement rule.
$P(\text{first red}) = \dfrac{6}{9}$ . Given the first was red, $8$ pens remain, $5$ of them red: $P(\text{second red}\mid\text{first red})=\dfrac{5}{8}$ .
$P(\text{both red}) = \dfrac{6}{9}\times\dfrac{5}{8} = \dfrac{30}{72} = \dfrac{5}{12}$ .
Apply the complement rule: $P(\text{at least one blue}) = 1-\dfrac{5}{12} = \dfrac{7}{12}$ .

Exam Tip

"At least one" questions are almost always faster by finding the probability of the opposite case (none at all) and subtracting from $1$ , rather than listing and adding every combination that includes at least one blue.