Two events are mutually exclusive when they cannot happen on the same trial. When that is true, the probability of “A or B” is simply P(A) + P(B). When it is not true, adding double-counts the shared outcomes.
This skill is the starting point of probability reasoning, and every tree diagram and total-probability check later in the module depends on it.
How do you represent outcomes so nothing is missed?
Start by listing every possible outcome of one trial. Then write each event as a set of those outcomes. Two events are mutually exclusive only if their sets share no outcome.
For a fair six-sided die the outcomes are 1, 2, 3, 4, 5, 6, each with probability 1/6. The event “a prime number” is {2, 3, 5}.
The event “a 6” is {6}. They share nothing, so they are mutually exclusive.
That gives P(prime or 6) = 3/6 + 1/6 = 4/6 = 2/3.
Listing first does two jobs. It shows you whether events overlap, and it gives you a way to check the answer by counting outcomes.
Worked example
A box holds 40 raffle tickets: 14 red, 10 blue, 12 green and 4 yellow. One ticket is drawn at random. Find (a) P(red or green), and (b) P(not blue).
Step 1, check the model. 14 + 10 + 12 + 4 = 40, so every ticket has a colour and each ticket has exactly one colour. The colours are mutually exclusive.
Step 2, write each probability. P(red) = 14/40, P(green) = 12/40, P(blue) = 10/40, P(yellow) = 4/40.
Step 3, (a). P(red or green) = 14/40 + 12/40 = 26/40 = 13/20.
Step 4, (b). P(not blue) = 1 − 10/40 = 30/40 = 3/4.
Check (b) a second way. Add the other three colours: 14 + 12 + 4 = 30, so 30/40 = 3/4. Both routes agree.
The mistake to watch for
The addition rule is applied to events that overlap.
Ten cards are numbered 1 to 10. Find P(even or a multiple of 3).
Mistaken working: P(even) = 5/10, P(multiple of 3) = 3/10, so the answer is 5/10 + 3/10 = 8/10.
The number 6 is both even and a multiple of 3, so it was counted twice.
Correction. List the outcomes. Even: 2, 4, 6, 8, 10. Multiples of 3: 3, 6, 9. Together the distinct outcomes are 2, 3, 4, 6, 8, 9, 10, which is 7 outcomes. The answer is 7/10.
The same result comes from 5/10 + 3/10 − 1/10 = 7/10, where the subtracted 1/10 is the shared outcome 6. Before you add, ask one question: “Can both events happen on the same trial?”
Check yourself
Try these, then open each answer.
1. A fair die is rolled. Find P(prime or 6).
Show answer
Prime: {2, 3, 5}. Six: {6}. No shared outcome, so add: 3/6 + 1/6 = 4/6 = 2/3.
2. A bag has red, green and white beads. P(red) = 0.3 and P(green) = 0.45. Find P(white) and P(red or green).
Show answer
Red, green and white are the only colours and cannot overlap. P(red or green) = 0.3 + 0.45 = 0.75. P(white) = 1 − 0.75 = 0.25.
3. A fair die is rolled. Are “odd number” and “greater than 4” mutually exclusive? Find P(odd or greater than 4).
Show answer
Odd: {1, 3, 5}. Greater than 4: {5, 6}. The number 5 is in both, so they are not mutually exclusive. Distinct outcomes: 1, 3, 5, 6, so P = 4/6 = 2/3. (Adding 3/6 + 2/6 = 5/6 would be wrong.)
Where this leads next
With outcomes listed cleanly, you are ready to use a two-stage probability tree, where each branch is one outcome of one stage. The non-calculator working trainer is useful for keeping fractions exact, and the percentage-base explorer helps when probabilities are given as percentages.
Students who follow each step here can still lose marks by skipping the listing step under time pressure. That is a pattern our teachers look for in online one-to-one Mathematics tuition.