How do you find the probability of at least one success?

Take 1 minus the probability of none. At least one is the exact opposite of none, so P(at least one) = 1 minus P(none). This is faster and safer than listing and adding every winning branch, which is slow and double counts the cases where more than one event happens. For two picks with P(gold) = 0.05 each and replacement, P(none) = 0.95 times 0.95 = 0.9025, so P(at least one gold) = 1 minus 0.9025 = 0.0975, not 0.05 + 0.05 = 0.10.

Is at least one the same as both?

No. Both is only the smallest corner of at least one. At least one of two events includes the first happening, the second happening, or both, so it is always larger than the chance of both. For two picks with P(gold) = 0.05, P(both gold) = 0.05 times 0.05 = 0.0025, while P(at least one gold) = 0.0975. Reading at least one as both is a common grade 7 to 9 mark-loser.

A biased dice has P(6) = 0.38 and is rolled 150 times. How many rolls are expected to not land on 6?

93. The question wants the not-6 rolls, which is the complement of the given event. Switch to the complement first: P(not 6) = 1 minus 0.38 = 0.62, then scale by the rolls: 0.62 times 150 = 93. Scaling the given 0.38 gives 0.38 times 150 = 57, which is the expected number of sixes, the wrong event. As a check, 57 + 93 = 150.

GCSE Maths Higher · AQA · Combined & conditional probability

At least one is one minus none

For an at least one problem, do not chase and add the winning branches. At least one is the exact opposite of none, so take 1 minus the chance it never happens: $P(\text{at least one}) = 1 - P(\text{none})$ . Adding the winning branches is slow and double counts the cases where more than one event occurs, and it is the single biggest grade 7 to 9 mark-loser in at-least-one questions.

The other half of the trap is reading at least one as both. Getting both is only the smallest corner of at least one, so the two answers are different. For two picks with $P(\text{gold}) = 0.05$ , $P(\text{both}) = 0.0025$ but $P(\text{at least one}) = 0.0975$ . At least one is always larger than both.

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How to spot it in your own work

You added the winning branches for an at least one question instead of taking $1 - P(\text{none})$ .
You read at least one as both, so your answer came out far smaller than it should be.
You scaled the given probability when the question asked about its opposite, the complement.
Your at-least-one answer was smaller than the chance of just one of the events, which is impossible.

An exam question that triggers it

Here is the structure of JUN24 Paper 3 Question 12b:

A bag contains a gold coin with probability 0.05.

A coin is picked, looked at, then replaced, and a second coin is picked, so each pick has $P(\text{gold}) = 0.05$ .

Find the probability of at least one gold.

The misconception adds the branches, $0.05 + 0.05 = 0.10$ , double counting the both-gold corner. The fix: use the complement, $P(\text{none}) = 0.95^2 = 0.9025$ , so $P(\text{at least one}) = 1 - 0.9025 = 0.0975$ .

Why students fall for this

At least one feels like a sum. The phrase invites students to find each way it can happen and add the chances, which works in principle but is slow and double counts the overlap where more than one event occurs. The complement route sidesteps all of that with a single subtraction.

None is easy to miss as the hook. Students do not see that the opposite of at least one is a single, clean and-branch, no event at all, whose probability is quick to find and then subtract from 1.

At least one and both blur together. Under time pressure at least one gets read as both, and because both is only the smallest corner, the answer collapses to a value far too small without any sense-check that would catch it.

The fix: Flip to the complement, then subtract from one

Step 1: name the opposite. The opposite of at least one is none. The opposite of fail at least one is pass both. The opposite event is a single clean outcome.

Step 2: find P(none). None is an and along a branch, so multiply the not-happening chances: $P(\text{none}) = 0.95 \times 0.95 = 0.9025$ .

Step 3: subtract from one. $P(\text{at least one}) = 1 - P(\text{none}) = 1 - 0.9025 = 0.0975$ .

Step 4: sense-check against both. At least one must be larger than both, since both is only one corner. If your answer is smaller than the chance of a single event, you read at least one as both.

Worked example

JUN24·3·12b structure: two picks with $P(\text{gold}) = 0.05$ and replacement. Find P(at least one gold).

Find P(none).
$P(\text{none}) = 0.95 \times 0.95 = 0.9025$
Subtract from one. $P(\text{at least one}) = 1 - 0.9025 = 0.0975$ . Trap: $0.05 + 0.05 = 0.10$ , which double counts both gold.

JUN24·3·10 cousin: a biased dice with $P(6) = 0.38$ rolled 150 times. Find the expected number of not-6 rolls.

Switch to the complement. $P(\text{not } 6) = 1 - 0.38 = 0.62$ .
Scale by the rolls. $0.62 \times 150 = 93$ . Trap: scaling the given chance gives $0.38 \times 150 = 57$ , the expected sixes.

NOV24·3·14b structure: a conditional test with $P(\text{pass A}) = 0.85$ and $P(\text{pass B} \mid \text{pass A}) = 0.78$ . Of 5000 entrants, 40% sit it. Find the expected number who fail at least one section.

Use the complement. $P(\text{pass both}) = 0.85 \times 0.78 = 0.663$ , so $P(\text{fail at least one}) = 1 - 0.663 = 0.337$ .
Scale by who sits. 40% of 5000 is 2000, so $0.337 \times 2000 = 674$ . Trap: reading at least one as both gives a far smaller figure, around 108.

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Common questions

How do you find the probability of at least one success?: Take $1 - P(\text{none})$ . At least one is the opposite of none, so one subtraction captures every winning case at once. For two picks with $P(\text{gold}) = 0.05$ , $P(\text{none}) = 0.9025$ , so $P(\text{at least one}) = 0.0975$ , not 0.10.
Is at least one the same as both?: No. Both is only the smallest corner of at least one, so at least one is always larger. Here $P(\text{both}) = 0.0025$ while $P(\text{at least one}) = 0.0975$ .
A biased dice has $P(6) = 0.38$ over 150 rolls. How many are expected to not land on 6?: 93. Switch to the complement, $P(\text{not } 6) = 0.62$ , then scale: $0.62 \times 150 = 93$ . Scaling the given 0.38 gives 57, the expected sixes, the wrong event.

Related misconceptions

AND events multiply along the branchAlong a branch, and means multiply, never add, so the combined chance is smaller than either step. The same multiply gives P(none) for the complement route.
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.

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