Topics

Probability trees and Venn diagrams

Covers AS91267

What a tree diagram is for

A tree diagram maps out events that happen in stages — two coin tosses, two counters drawn from a bag, a test followed by a re-test.
Each branch is one possible outcome of that stage, labelled with its probability.
The branches leaving any one point always add to 1, because something must happen. This is the fastest way to check you have not missed a branch or mislabelled one.
Each complete path from left to right is one overall outcome, and every path together covers every possibility exactly once.

The two rules: multiply along, add between

Multiply ALONG a path — this and then that.
- The stages happen one after another, so both must occur for that path to be the outcome.
- A path's probability is the product of every branch on it.
ADD separate paths — this path or that path.
- Different paths are different ways of succeeding, and only one of them can actually happen.
A quick way to keep them straight: "and" narrows things down, so the number gets smaller (multiplying); "or" gives you more ways to win, so the number gets bigger (adding).
"At least one" is almost always easier backwards. There are usually many paths with at least one success but only one path with none at all:

$P(\text{at least one}) = 1 - P(\text{none})$

With and without replacement

With replacement — the item goes back in, so the bag is identical for the second draw.
- The totals are unchanged, so every stage has the same probabilities.
- The stages are independent: the first result tells you nothing about the second.
Without replacement — the item is kept out, so the bag has changed.
- The total drops by one, and so does the count of whatever was taken.
- The second-stage probabilities are different on every branch, because what is left depends on what was drawn first.
- The stages are dependent — the second-stage branches are conditional probabilities.
Decide which one applies before drawing anything. The phrase to hunt for is "replaced" or "not replaced"; failing that, ask whether the item could physically be picked twice.

Venn diagrams

A Venn diagram shows events as overlapping circles inside a rectangle that represents every possible outcome.
Reading the regions:
- the overlap (intersection) is " $A$ and $B$ " — outcomes in both
- everything inside either circle is " $A$ or $B$ " (the union) — including the overlap
- the space outside both circles is "neither", which is easy to forget and often carries a mark
"Or" in mathematics includes both. Everyday English often means one or the other but not both; a Venn diagram makes the difference visible.

Mutually exclusive is not the same as independent

Mutually exclusive events cannot both happen:
- their circles do not overlap at all
- $P(A \text{ and } B) = 0$
- example: a single card is a heart or a spade, never both
Independent events do not affect each other's chances:
- $P(A \text{ and } B) = P(A) \times P(B)$
- example: two separate coin tosses
These two ideas are opposites, not synonyms — and confusing them is the most common conceptual error in the topic.
- Mutually exclusive events are strongly dependent: if one happens, the other definitely cannot, so knowing about one tells you a great deal about the other.
- Independent events with non-zero probabilities must overlap, because they can both happen at once.

Worked ExampleMultiplying along a path

A fair coin is tossed twice. Find the probability of getting two heads.

Step 1 — Probability of heads each toss

Each toss is independent with $P(\text{heads}) = \tfrac{1}{2}$ .

Step 2 — Multiply along the H–H path

"Heads and heads" means multiply along the single path:

$P(HH) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

Worked ExampleDrawing without replacement

A bag has 4 red and 6 blue counters. Two are drawn without replacement. Find the probability that both are red.

Step 1 — First draw

$P(\text{red}) = \frac{4}{10}$

Step 2 — Second draw (one red already gone)

Now 3 red remain out of 9 counters:

$P(\text{red} \mid \text{first red}) = \frac{3}{9}$

Step 3 — Multiply along the path

$P(\text{both red}) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$

Worked Example"At least one" — using the complement

The same bag holds 4 red and 6 blue counters. Two counters are drawn without replacement.

Find the probability that at least one is red.

Step 1 — List what "at least one red" actually covers

Three of the four paths qualify: red-red, red-blue, and blue-red. Only blue-blue does not.

Step 2 — Turn it around, because one path is quicker than three

Everything except blue–blue counts, so:

$P(\text{at least one red}) = 1 - P(\text{no red}) = 1 - P(\text{blue, blue})$

Step 3 — Find the single unwanted path

First draw: 6 blue out of 10. Second draw: one blue is gone, so 5 blue out of 9 remain:

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

Step 4 — Subtract from 1

$P(\text{at least one red}) = 1 - \frac{1}{3} = \frac{2}{3}$

$P(\text{at least one red}) = \frac{2}{3} \approx 0.667$

Tips and Tricks

Draw the tree even when the question doesn't ask for one. Method marks live in the labelled branches, and a tree makes a missing path obvious in a way that a line of algebra never does.

"At least one" means use the complement: $P(\text{at least one}) = 1 - P(\text{none})$ . "None" is a single path; the alternative is adding every other path.

Check each set of branches adds to 1 before going any further. If a pair reads $\tfrac{4}{10}$ and $\tfrac{5}{10}$ , a branch is wrong and everything downstream will be too.

Without replacement, both numbers change on the second draw — the count you want and the total (from $\tfrac{4}{10}$ to $\tfrac{3}{9}$ ). Dropping only the total is the most common tree-diagram error.

Do not say "mutually exclusive" when you mean "independent". Mutually exclusive events cannot happen together; independent events do not influence each other. Using the wrong term loses the mark even when the arithmetic is right.