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Two events are mutually exclusive if they cannot occur at the same time (i.e. they have no outcomes in common).
Dua peristiwa disebut mutually exclusive jika mereka tidak dapat terjadi pada saat yang sama (yaitu mereka tidak memiliki hasil yang sama).
Mutually Exclusive Events

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Addition Rule 1

When two events A and B are mutually exclusive, the probability that A or B will occur is
     P(A or B)=P(A)+P(B)~~~~~ P(A \text{ or } B) = P(A) + P(B)

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Addition Rule 1 - Example

At a political rally, there are 20 Republicans (R), 13 Democrats (D), and 6 Independents (I). If a person is selected, find the probability that he or she is either a Democrat or an Independent.
Solution:P(D or I)=P(D)+P(I)=1339+639=1939P(D \text{ or } I) = P(D) + P(I) = \dfrac{13}{39} + \dfrac{6}{39} = \dfrac{19}{39}

A day of the week is selected at random. Find the probability that it is a weekend.
Solution:P(Saturday or Sunday)=P(Saturday)+P(Sunday)=17+17=27P(\text{Saturday or Sunday}) = P(\text{Saturday}) + P(\text{Sunday}) = \dfrac{1}{7} + \dfrac{1}{7} = \dfrac{2}{7}

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Addition Rule 2

When two events A and B are not mutually exclusive, the probability that A or B will occur is
     P(A or B)=P(A)+P(B)−P(A and B)~~~~~ P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)
Addition Rule 2

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Addition Rule 2 - Example

In a hospital unit there are eight nurses and five physicians. Seven nurses and three physicians are females. If a staff person is selected, find the probability that the subject is a nurse or a male.
STAFFFEMALESMALESTOTAL
NURSE718
PHYSICIAN325
TOTAL10313

Solution:P(nurse or male)=P(nurse)+P(male)+P(male and nurse)=813+313−113=1013P(\text{nurse or male}) = P(\text{nurse}) + P(\text{male}) + P(\text{male and nurse}) = \dfrac{8}{13} + \dfrac{3}{13} - \dfrac{1}{13} = \dfrac{10}{13}

On New Year’s Eve, the probability that a person driving while intoxicated is 0.32, the probability of a person having a driving accident is 0.09, and the probability of a person having a driving accident while intoxicated is 0.06. What is the probability of a person driving while intoxicated or having a driving accident?
Solution:P(intoxicated or accident)=P(intoxicated)+P(accident)−P(intoxicated and accident)=0.32+0.09−0.06=0.35P(\text{intoxicated or accident}) = P(\text{intoxicated}) + P(\text{accident}) - P(\text{intoxicated and accident}) = 0.32 + 0.09 - 0.06 = 0.35