If we have two events, A and B, that we are interested in, we often want to know the probability that either A or B occurs. Note the use of the word "or," the key to this rule. The "or" is what we call an "inclusive or." In other words, either one event can occur or both events can occur.
Such probabilities are calculated using the addition rule for probabilities.
The logic behind this formula is that when P(A) and P(B) are added, the occasions on which A and B both occur are counted twice. To adjust for this, P(AB) is subtracted.
If events A and B are mutually exclusive, the joint probability of A and B is 0. Consequently, the probability that either A or B occurs is simply the sum of the unconditional probabilities of A and B: P (A or B) = P(A) + P(B).
What is the probability that a card selected from a deck will be either an ace or a spade? The relevant probabilities are:
P(ace) = 4/52; P(spade) = 13/52
The only way in which an ace and a spade can both be drawn is to draw the ace of spades. There is only one ace of spades, so:
P(ace and spade) = 1/52.
The probability of an ace or a spade can be computed as:
P(ace) + P(spade) - P(ace and spade) = 4/52 + 13/52 - 1/52 = 16/52 = 4/13.
Consider the probability of rolling dice twice and getting a 6 on at least one of the rolls. The events are defined in the following way:
Event A: 6 on the first roll: p(A) = 1/6
Event B: 6 on the second roll: p(B) = 1/6
P(A and B) = 1/6 x 1/6
P(A or B) = 1/6 + 1/6 - 1/6 x 1/6 = 11/36
The same answer can be computed using the following (admittedly convoluted) approach: Getting a 6 on either roll is the same thing as not getting a number from 1 to 5 on both rolls. This is equal to: 1 - P(1 to 5 on both rolls).
The probability of getting a number from 1 to 5 on the first roll is 5/6. Likewise, the probability of getting a number from 1 to 5 on the second roll is 5/6. Therefore, the probability of getting a number from 1 to 5 on both rolls is: 5/6 x 5/6 = 25/36. This means that the probability of not getting a 1 to 5 on both rolls (getting a 6 on at least one roll) is: 1-25/36 = 11/36.
Despite the convoluted nature of this method, it has the advantage of being easy to generalize to three or more events. For example, the probability of rolling dice three times and getting a six on at least one of the three rolls is: 1 - 5/6 x 5/6 x 5/6 = 0.421