The probability of an event A is P(A) = number of favourable outcomes ÷ total outcomes in the sample space, always a value from 0 to 1. List the sample space first so you count correctly, then apply the complement rule P(not A) = 1 − P(A) whenever a question asks for "not", "at least one", or anything easier to find by working out its opposite.
Probability questions look simple, so students rush the first step, listing the sample space, and get it wrong before any calculation happens. The most common error is forgetting that the outcomes you divide by must be equally likely. Rolling two dice and asking for the total is the classic trap: there are 11 possible totals (2 to 12) but they are not equally likely, so the sample space is the 36 equally likely ordered pairs, giving P(total of 7) = 6/36 = 1/6, not 1/11.
A second common error is not spotting when the complement rule would save time. Questions phrased as "at least one" or "not" are often much faster to answer by finding the opposite event and subtracting from 1, rather than listing every case directly.
Step 1
List every outcome (the sample space)
Step 2
Count outcomes matching the event
Step 3
P(A) = favourable ÷ total
Step 4
P(not A) = 1 − P(A)
A bag contains 5 red balls, 3 blue balls, and 2 green balls. A ball is drawn at random. Find the probability that it is blue.
Solution
3 ÷ 10 = 3/10.Quick check: every probability must lie between 0 and 1 inclusive. 3/10 = 0.3, which is a sensible answer. If your working gives a fraction bigger than 1, you have counted something wrong.
A fair six-sided die is rolled once. Find the probability of NOT rolling a 5.
Solution
1/6.P(not 5) = 1 − 1/6 = 5/6.A box has 4 defective and 16 working light bulbs. One bulb is picked at random. Find the probability that it is working.
Solution
P(defective) = 4/20 = 1/5.1 − 1/5 = 4/5. This matches counting directly (16 ÷ 20 = 4/5), confirming the answer.The step students get wrong
Students count the favourable outcomes carefully, then divide by the wrong total. Two usual causes: the listed outcomes are not equally likely, or an item has been drawn and not replaced, so the total drops for the next draw (10 balls become 9, not 10). Before writing P(A) = favourable ÷ total, check that every outcome in your list is just as likely as every other one, and re-read the question to see whether the sample space changes between draws.
Use P(not A) = 1 − P(A) whenever the direct count is harder than the opposite count. This is especially useful for "at least one" questions involving several trials, where listing every success case is slow but listing the single failure case (zero successes) is quick. Tree diagrams often combine with the complement rule for these multi-trial questions.
Exam habit: before listing outcomes, ask "is it easier to count what I want, or what I don't want?" If the "not" version has fewer cases, use the complement rule and subtract from 1.
What is the difference between an outcome and an event?
An outcome is a single result of a trial, such as rolling a 4. An event is a set of one or more outcomes that satisfy a condition, such as "rolling an even number", which covers the outcomes 2, 4, and 6. Probability is always calculated for an event, using the outcomes that belong to it.
Can probability ever be negative or greater than 1?
No. Probability always lies between 0 (impossible) and 1 (certain) inclusive. If a calculation gives a value outside this range, there is an error, usually in counting the sample space or the favourable outcomes.
How is this different from mutually exclusive or independent events?
This guide covers a single event from one sample space. Once a question involves two or more events, whether they can happen together (mutually exclusive) or whether one affects the other (independent) changes how you combine their probabilities. See mutually exclusive vs independent events for that next step. Questions combining several overlapping events are often easier to picture with a Venn diagram and set notation before you calculate anything.
When do I need a tree diagram instead of this method?
Once a question involves two or more trials in sequence, such as drawing two balls one after another, a tree diagram keeps the branches and combined probabilities organised, especially when the sample space changes between draws.
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