Mutually Exclusive Events

Events that cannot both happen

Roll a fair dice once. The event "rolls a 1" and the event "rolls a 6" cannot both happen: one roll shows one face. Two events that cannot happen at the same time are called mutually exclusive.

On a Venn diagram, the overlap of two events holds the outcomes in both. For mutually exclusive events there are no such outcomes, so the overlap holds 0.

Count the faces

Put all 6 faces in the diagram. The face 1 goes in the circle "rolls a 1", the face 6 goes in the circle "rolls a 6", and nothing goes in the overlap. The other 4 faces, 2, 3, 4 and 5, are in neither event, so they sit in the box outside both circles. Check: 1 + 0 + 1 + 4 = 6.

One face in each circle, 0 in the overlap, and 4 faces outside both.

Add the chances

The event "rolls a 1 or a 6" is everything inside either circle: 2 of the 6 faces, so its probability is $\frac{2}{6}=\frac{1}{3}$2/6 = 1/3. The same answer comes from adding the two chances: $\frac{1}{6}+\frac{1}{6}=\frac{2}{6}$1/6 + 1/6 = 2/6.

Adding works because no face is in both circles, so no face is counted twice. For mutually exclusive events A and B, P(A or B) = P(A) + P(B).

Either circle: 1 + 0 + 1 = 2 of the 6 faces, which is $\frac{1}{6}+\frac{1}{6}$1/6 + 1/6.

When events overlap

Now take "rolls an even number", the faces 2, 4 and 6, and "rolls more than 3", the faces 4, 5 and 6. These can happen together: a roll of 4 or 6 is in both, so the events are not mutually exclusive.

Adding the chances gives $\frac{3}{6}+\frac{3}{6}=1$3/6 + 3/6 = 1, which says a roll is certain to be even or more than 3. It is not: a 1 or a 3 is neither. Only 4 faces, 2, 4, 5 and 6, are in either event, so the chance is $\frac{4}{6}$4/6. Adding counted the faces 4 and 6 twice, and the addition rule takes the overlap off once: $\frac{3}{6}+\frac{3}{6}-\frac{2}{6}=\frac{4}{6}$3/6 + 3/6 − 2/6 = 4/6.

Even and more than 3 share the faces 4 and 6, so the overlap holds 2, and adding the two circles counts those faces twice.

Outcomes that cover every case

A spinner lands on 1, 2 or 3, and on exactly one of them. The three outcomes are mutually exclusive, and together they cover every case, so their probabilities add to 1. If 1 has probability $\frac{1}{8}$1/8 and 2 has $\frac{3}{8}$3/8, then 3 has $1-\frac{1}{8}-\frac{3}{8}=\frac{4}{8}=\frac{1}{2}$1 − 1/8 − 3/8 = 4/8 = 1/2.

Mutually exclusive events do not have to cover every case. Rolling a 1 and rolling a 6 are mutually exclusive, but their chances add to only $\frac{2}{6}$2/6, because 4 faces are in neither event. They are not complements: the complement of rolling a 1 is rolling 2, 3, 4, 5 or 6.

The section for 1 takes $\frac{1}{8}$1/8 of the spinner, 2 takes $\frac{3}{8}$3/8 and 3 takes the other $\frac{4}{8}$4/8. The three fill the whole circle, 1.

Not the same as independent

Two mutually exclusive events that can each happen are not independent. If a roll shows a 1, it certainly does not show a 6, so knowing one event happened changes the chance of the other to 0. For mutually exclusive events P(A and B) = 0; it is independent events, a separate idea, whose chances multiply.

The usual mistakes

Multiplying the chances, $\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}$1/6 × 1/6 = 1/36. Multiplying gives the chance of both happening, which for mutually exclusive events is 0; for "one or the other", add.

Adding the denominators as well as the numerators. Both chances are already out of 6, so $\frac{1}{6}+\frac{1}{6}=\frac{2}{6}$1/6 + 1/6 = 2/6, not $\frac{2}{12}$2/12.

Adding the chances of events that can overlap. Even and more than 3 share two faces, so their chances do not simply add.

Taking a missing outcome as 1 minus only one of the given chances. All the other chances must come off 1.

A raffle

In the application below, the prize for a multiple of 3 or a multiple of 5 overlaps at 15 and 30, so the overlap is taken off once. A second prize goes only to tickets that did not win the first, so the two prizes are mutually exclusive and their chances add.