Do you add or multiply probabilities for two events that both happen?

You multiply. Along a branch of a probability tree, and means multiply, because each stage keeps only a fraction of what survived the stage before, so the combined chance is smaller than either part. P(Yes from Bag A and Yes from Bag B) with chances 3/5 and 1/10 is 3/5 times 1/10 = 3/50 = 0.06, not 3/5 + 1/10 = 7/10. Adding gives an answer larger than either single chance, which is impossible for a both-events probability.

Bag A has a green ball with probability 1/5 and Bag B with probability 3/10. What is the probability that both are green?

3/50, which is 0.06. Multiply along the branch: 1/5 times 3/10 = 3/50. The trap, 1/5 + 3/10 = 1/2, adds the chances and gives an answer larger than the 3/10 single chance, which a both probability can never be. The quick self-check is that a both probability is always smaller than either branch.

How do you handle the same event happening n times, like (1/2) to the power n = 1/64?

A repeated and is a power, not a multiply by the count. Each extra repeat multiplies the probability by another 1/2, so the count sits in the exponent. Write 1/64 as a power of 2: since 2 to the power 6 = 64, 1/64 = (1/2) to the power 6, so n = 6. The trap treats the repeats linearly, as 1/2 times n or 64/2, and answers n = 32.

GCSE Maths Higher · AQA · Combined & conditional probability

AND events multiply along the branch

Along a branch of a probability tree, and means multiply, never add. Each stage keeps only a fraction of what survived the stage before, so the combined chance of two events is smaller than either part. For independent events $P(A \text{ and } B) = P(A) \times P(B)$ , and for dependent ones $P(A) \times P(B \mid A)$ . Adding the branch probabilities is the single biggest grade 7 to 9 mark-loser in tree questions.

The other half of the trap is the repeated event. The same event happening n times is that fraction raised to a power, $p^n$ , not the fraction multiplied by how many times. Three Heads in a row is $(1/2)^3 = 1/8$ , not $1/2 \times 3$ . The count belongs in the exponent.

Already know this is your gap? Skip the diagnostic and jump straight into the targeted lesson.

How to spot it in your own work

You added the branch probabilities for an and, when you should have multiplied them.
Your both-events answer came out larger than one of the single branch probabilities, which is impossible.
For a repeated event you wrote $p \times n$ instead of $p^n$ .
You found an expected frequency by adding the chances first, then scaling by the number of trials.

An exam question that triggers it

Here is the structure of JUN22 Paper 1 Question 12b:

A game uses two bags.

$P(\text{Yes from Bag A}) = \tfrac{3}{5}$ and $P(\text{Yes from Bag B}) = \tfrac{1}{10}$ .

A player wins only with a Yes on both bags. The game is played 450 times. Find the expected number of winners.

The misconception adds first, $\tfrac{3}{5} + \tfrac{1}{10} = \tfrac{7}{10}$ , then scales to $\tfrac{7}{10} \times 450 = 315$ . The fix: multiply first, $\tfrac{3}{5} \times \tfrac{1}{10} = \tfrac{3}{50}$ , then $\tfrac{3}{50} \times 450 = 27$ .

Why students fall for this

Adding is met first and overlearned. Students see "this or that" problems where probabilities add, and the habit carries over to "this and that" without the switch to multiplying. The word and reads like a signal to combine, and combining feels like adding.

The shrink is counter-intuitive. It feels as though wanting two things should be a bigger demand and so a bigger number, when in fact a both-events chance is always smaller than either part. Students who add never run the sense-check that would catch the answer exceeding a single branch.

Repeats hide the power. A question asks for the same event n times, and a half-recalled rule reaches for $p \times n$ because that is the simpler operation, missing that each extra and is another factor and so an exponent.

The fix: Walk one branch, multiply, then sense-check

Step 1: identify the and. Wanting one event and another, along a single branch of the tree, is a multiply, not an add.

Step 2: multiply the branch probabilities. $P(A \text{ and } B) = P(A) \times P(B)$ for independent events, or $P(A) \times P(B \mid A)$ when the second depends on the first.

Step 3: sense-check against the branches. A both-events answer must be smaller than every single probability it was built from. If it is larger, you added by mistake.

Step 4: for repeats, use a power. The same event n times is $p^n$ . To solve $(1/2)^n = 1/64$ , write the target as a matching power and equate exponents.

Worked example

NOV23·3·10b structure: Bag A green with probability $\tfrac{1}{5}$ , Bag B green with probability $\tfrac{3}{10}$ . Find P(both green).

And means multiply.
$\tfrac{1}{5} \times \tfrac{3}{10} = \tfrac{3}{50} = 0.06$
Sense-check. $0.06$ is smaller than both $\tfrac{1}{5}$ and $\tfrac{3}{10}$ , as a both probability must be. Trap: $\tfrac{1}{5} + \tfrac{3}{10} = \tfrac{1}{2}$ , larger than $\tfrac{3}{10}$ .

JUN22·1·12b structure: two bags with $P(\text{Yes}) = \tfrac{3}{5}$ and $\tfrac{1}{10}$ , win on both, 450 plays. Find the expected winners.

Multiply for the and. $\tfrac{3}{5} \times \tfrac{1}{10} = \tfrac{3}{50}$ .
Scale by the trials. $\tfrac{3}{50} \times 450 = 27$ . Trap: adding first gives $\tfrac{7}{10} \times 450 = 315$ .

NOV23·2·11 structure: a fair event with probability $\tfrac{1}{2}$ repeated n times, $(1/2)^n = 1/64$ . Find n.

Write the target as a power. $2^6 = 64$ , so $\tfrac{1}{64} = (1/2)^6$ .
Equate exponents. $n = 6$ . Trap: treating the repeats linearly gives $n = 32$ .

Find out if this is costing you marks

The 10-minute diagnostic checks for this pattern (and four others) using AQA-style GCSE Higher items. Free, no signup, anonymous.

Take the diagnostic →

Common questions

Do you add or multiply for two events that both happen?: Multiply. Along a branch, and means multiply, so $P(A \text{ and } B) = P(A) \times P(B)$ , and the answer is smaller than either part. Adding gives a value larger than a single branch, which is impossible.
Two bags give green with chances $\tfrac{1}{5}$ and $\tfrac{3}{10}$ . What is P(both green)?: $\tfrac{1}{5} \times \tfrac{3}{10} = \tfrac{3}{50} = 0.06$ . The trap, $\tfrac{1}{2}$ , adds the chances and exceeds the $\tfrac{3}{10}$ branch.
How do you solve $(1/2)^n = 1/64$ ?: A repeated and is a power, so the count sits in the exponent. Since $2^6 = 64$ , $\tfrac{1}{64} = (1/2)^6$ , giving $n = 6$ , not the linear answer 32.

Related misconceptions

At least one uses the complement ruleP(at least one) = 1 minus P(none at all). The complement route is the standard Higher method and avoids double-counting the cases where more than one event occurs.
Conditional denominator is the restricted setWhen a condition is stated or a draw is made without replacement, the denominator shrinks to the remaining or matching set, not the original total.

← All GCSE Maths Higher misconceptions