A spinner lands on red with probability 0.3. What is P(red both times) in two spins?

0.09. 'And' multiplies, so P(red then red) = 0.3 × 0.3 = 0.09. Needing two things to both happen is harder than needing one, so the answer must be smaller than 0.3 — and 0.09 is. The common wrong answer 0.6 adds the probabilities (0.3 + 0.3); that is bigger than a single spin, which is the tell that adding is wrong.

Two independent events have probabilities 0.8 and 0.4. What is the probability both happen?

0.32. For independent events you multiply: 0.8 × 0.4 = 0.32. Adding gives 0.8 + 0.4 = 1.2, which is impossible because a probability cannot exceed 1 — so an additive answer above 1 is wrong on sight.

What probability goes on the other branch of a probability tree?

1 − p. The two branches out of any point on a tree must add to 1, because they cover everything that can happen there. If 'pass' has probability 0.4, the 'not pass' branch is 1 − 0.4 = 0.6. You never leave it blank or copy the other branch. On a tree you multiply along a branch and add between completed paths.

GCSE Maths Foundation · AQA · Probability

Combined event probability: why 'and' multiplies, not adds

Combined-event questions trip students who find the chance of two things both happening by adding the probabilities. Told a spinner lands on red with probability 0.3 and spun twice, they answer 0.6 — $0.3 + 0.3$ — instead of $0.3 \times 0.3 = 0.09$ . But two reds is rarer than one, so the answer must be smaller than 0.3, not bigger.

The thirty-second fix: “and” multiplies. For two independent events both happening you multiply the probabilities, and the answer comes out smaller than either one. On a tree you multiply along a branch, add between completed paths, and the other branch is 1 − p.

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How to spot it in your own work

You found P(A and B) by adding — writing $0.3 + 0.3 = 0.6$ instead of $0.3 \times 0.3 = 0.09$ .
Your “and” answer came out bigger than each single event, when it should be smaller.
You wrote a probability above 1 — such as $0.8 + 0.4 = 1.2$ — without noticing that is impossible.
You left the other tree branch blank, or copied it, instead of using $1 - p$ (so the pair sums to 1).
You labelled a tree branch with a frequency (a count like 10) rather than a probability (like $0.25$ ).

An exam question that triggers it

Here is the canonical AQA Foundation trigger (Jun24 P2 Q18 / Jun22 P2 Q20 shape, calculator):

Jose passes his test with probability 0.8. Maria passes hers with probability 0.4. The events are independent. Work out the probability that they both pass.

The misconception answer is $0.8 + 0.4 = 1.2$ — adding the probabilities. But 1.2 is greater than 1, so it cannot be a probability at all.

“And” multiplies: $0.8 \times 0.4 = 0.32$ .

Why students fall for this

Addition is the first operation students reach for when they see two numbers and the word “and”. But needing two independent things to both happen is a fraction of a fraction: of the 0.3 of the time red comes up first, only 0.3 of those are followed by another red. That is multiplication, and multiplying two numbers below 1 always makes the result smaller.

The same belief shows up on probability trees. Students treat the two branches out of a point as independent labels that need not relate, so a “not pass” branch is left blank or copied. In fact the branches out of a point are everything that can happen there, so they must sum to 1: the other branch is $1 - p$ . Some students even write a frequency (a count) on a branch instead of a probability.

AQA Foundation calculator papers exploit every face of this: P(both pass) on two independent events (Jun22 P2 Q20), a two-stage tree for P(one red and one white) (Jun24 P2 Q18b), and the frequencies-versus-probabilities slip on the same tree (Jun24 P2 Q18a).

The fix: 'And' multiplies: multiply along a branch, add between paths, and the other branch is 1 − p

Two independent events both happening: multiply. P(red both times) = $0.3 \times 0.3 = 0.09$ . The answer is smaller than 0.3, because two reds is rarer than one. Adding (0.6) is bigger — the tell that it is wrong.

The other branch is 1 − p. If “pass” is 0.4, then “not pass” is $1 - 0.4 = 0.6$ . The branches out of a point sum to 1 — never leave one blank.

On a tree: multiply along, add between. You multiply the probabilities along a single path, and you add only when combining two separate completed paths that both count. And any probability above 1 is impossible, so reject it on sight.

Worked example

Spinner A wins with probability 0.7 and Spinner B wins with probability 0.4, independently. Work out the probability that exactly one of the two spinners wins.

Fill the other branches with 1 − p. A loses with $1 - 0.7 = 0.3$ ; B loses with $1 - 0.4 = 0.6$ .
Multiply along each winning-one path.
$P(\text{A win, B lose}) = 0.7 \times 0.6 = 0.42$
$P(\text{A lose, B win}) = 0.3 \times 0.4 = 0.12$
Add the two separate paths.
$0.42 + 0.12 = 0.54$

Check the whole tree adds to 1: $0.28 \text{ (both win)} + 0.54 \text{ (exactly one)} + 0.18 \text{ (both lose)} = 1.00$ . The trap is to add along a path — but $0.7 + 0.6 = 1.3$ exceeds 1, so it cannot be a probability. Multiply along a branch, add only between completed paths.

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Common questions

Why can’t I just add the two probabilities for “and”?: Because needing two things to both happen is harder than needing one, so the answer must be smaller, not bigger. Adding makes the number grow: $0.3 + 0.3 = 0.6$ is bigger than a single spin. Multiplying shrinks it correctly: $0.3 \times 0.3 = 0.09$ . If an “and” answer comes out above 1, it cannot be a probability.
When do I add probabilities, then?: You add when you combine two separate completed routes that each count as a win — for example P(exactly one wins) = P(A win, B lose) + P(A lose, B win). On a tree: multiply along a single path, add between different paths.
What goes on the other branch of a tree?: $1 - p$ . The branches out of one point cover everything that can happen there, so they sum to 1. If one branch is 0.4, the other is $1 - 0.4 = 0.6$ — never blank, never a copy.
Do I write a probability or a frequency on a tree branch?: A probability. A branch carries a probability such as $0.25$ , not a count such as 10. Expected frequencies (like “10 greens in 40 draws”) are a separate calculation — $P \times \text{number of trials}$ — done after the tree, not on it.

Related misconceptions

Reading Venn diagrams: why a labelled region is not the whole setThe neighbouring probability misconception — combining the parts of a diagram correctly, and using the right denominator for a probability.
Relative frequency and expected outcomes: why 'estimate' means P × trials, not a round numberThe experimental-probability companion — turning a probability into an expected count, and judging which estimate is most reliable.

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