Mutually exclusive events
In logic and probability theory, two propositions (or events) are mutually exclusive or disjoint if they cannot both be true (occur). A clear example is the set of outcomes of a single coin toss, which can result in either heads or tails, but not both.
In the coin-tossing example, both outcomes are, in theory, jointly exhaustive, which means that at least one of the outcomes must happen, so these two possibilities together exhaust all the possibilities. However, not all mutually exclusive events are collectively exhaustive. For example, the outcomes 1 and 4 of a single roll of a six-sided die are mutually exclusive (both cannot happen at the same time) but not collectively exhaustive (there are other possible outcomes; 2,3,5,6).
In logic, two mutually exclusive propositions are propositions that logically cannot be true in the same sense at the same time. To say that more than two propositions are mutually exclusive, depending on context, means that one cannot be true if the other one is true, or at least one of them cannot be true. The term pairwise mutually exclusive always means that two of them cannot be true simultaneously.
In probability theory, events E1, E2, ..., En are said to be mutually exclusive if the occurrence of any one of them implies the non-occurrence of the remaining n − 1 events. Therefore, two mutually exclusive events cannot both occur. Formally said, the intersection of each two of them is empty (the null event): A ∩ B = ∅. In consequence, mutually exclusive events have the property: P(A ∩ B) = 0.
For example, it is impossible to draw a card that is both red and a club because clubs are always black. If just one card is drawn from the deck, either a red card (heart or diamond) or a black card (club or spade) will be drawn. When A and B are mutually exclusive, P(A ∪ B) = P(A) + P(B). To find the probability of drawing a red card or a club, for example, add together the probability of drawing a red card and the probability of drawing a club. In a standard 52-card deck, there are twenty-six red cards and thirteen clubs: 26/52 + 13/52 = 39/52 or 3/4.
One would have to draw at least two cards in order to draw both a red card and a club. The probability of doing so in two draws depends on whether the first card drawn were replaced before the second drawing, since without replacement there is one fewer card after the first card was drawn. The probabilities of the individual events (red, and club) are multiplied rather than added. The probability of drawing a red and a club in two drawings without replacement is then 26/52 × 13/51 × 2 = 676/2652, or 13/51. With replacement, the probability would be 26/52 × 13/52 × 2 = 676/2704, or 13/52.
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