Another type of probability deals with mutually exclusive events. What do we mean by mutually exclusive events?
And what does it mean for two events not to be mutually exclusive? Consider the following example of drawing cards: Example 8 What is the probability that a card selected from a deck will be either an ace or a spade? a)2/52 b)2/13 c)7/26 d)4/13 e)17/52 Solution First, identify events A and B and notice the "or" in between them. That means a greater probability than either A or B individually. Therefore, we expect the answer to be greater than 4/52(ace)=1/13 or 13/52(spade)= 1/4. Eliminate a and b. The tricky part of this question lies in the fact that when we figure probability, we are really just counting, and sometimes, we count twice. In this case we have counted the ace of spades twice. If you don't see this, consider what the 4 in 4/52 stands for: ace of hearts, ace of diamonds, ace of clubs, ace of spades. The 13 in 13/52 stands for all the spades: 1,2,3King, Ace(of spades). Therefore if we just combined the probabilities by the rule for P(A or B) = P(A) + P(B) we would be over counting. We have to subtract 1/52, the ace of spades that was counted twice. Our answer becomes 4/52 + 13/52 - 1/52 = 16/52 = 4/13. Another way to think about the question is to just count aces and spades; that is, use simple probability. There are 13 spades in a deck and 3 aces other than the ace of spades already included in the 13 spades. Therefore, there are 16 desired outcomes out of a total of 52 possible outcomes, or 16/52 = 4/13. In the above example, events A and B are not mutually exclusive. Figuring the probability for event A includes part of the probability of event B, and we must therefore subtract out this "over-counted" probability to get the correct answer.
The following example illustrates mutually exclusive events: Example 9 What is the probability that a card selected from a deck will be either an ace or a king? a)1/169 b)1/26 c)2/13 d)4/13 e)8/13
Solution The question asks for either an ace or a king. Since there are four kings and four aces in a deck, the probabilities for event A and event B are the same, 4/52 = 1/13. Our answer must be more than this, so eliminate a and b. Do kings and aces have anything to do with each other? Is there such a thing as an ace of kings or a king of aces? No, so we don't have to worry about having over-counted; the events are mutually exclusive. The probability is straightforward: P(A or B) = P(A) + P(B) = 1/13 + 1/13 = 2/13. C is correct. Again we could have used simple probability. Count the total number of kings and aces (4+4) and divide by the total number of cards in a deck: 8/52 = 2/13
A conditional probability is the probability of an event given that another event has occurred. Example 10 What is the probability that the total of two dice will be greater than 8 given that the first die is a 6?
Solution This can be computed by considering only outcomes for which the first die is a 6. Then, determine the proportion of these outcomes that total more than 8. All the possible outcomes for two dice are shown in the section on simple probability. There are 6 outcomes for which the first die is a 6: (6,1),(6,2),(6,3), (6,4),(6,5),(6,6), and of these, there are four that total more than 8. The probability of a total greater than 8 given that the first die is 6 is therefore 4/6 = 2/3. 1. Probability of A and B If A and B are independent, then the probability that events A and B both occur is p(A and B) = p(A) p(B). In other words, the probability of A and B both occurring is the product of the probability of A and the probability of B. What is the probability that a coin will come up with heads twice in a row? Two events must occur: a heads on the first toss and a heads on the second toss. Since the probability of each event is 1/2, the probability of both events is: 1/2 1/2 = 1/4. Now consider a similar problem: someone draws a card at random out of a deck, replaces it, and then draws another card at random. What is the probability that the first card is the ace of clubs and the second card is a club (any club)? Since there is only one ace of clubs in the deck, the probability of the first event is 1/52. Since 13/52 = 1/4 of the deck is composed of clubs, the probability of the second event is 1/4. Therefore, the probability of both events is 1/52 1/4 = 1/208. What's the probability of A and B (2 of 2) if A and B are not independent? If A and B are not independent, then the probability of A and B is p(A and B) = p(A) p(B|A) where p(B|A) is the conditional probability of B given A. If someone draws a card at random from a deck and then, without replacing the first card, draws a second card, what is the probability that both cards will be aces? Event A is that the first card is an ace. Since 4 of the 52 cards are aces, p(A) = 4/52 = 1/13. Given that the first card is an ace, what is the probability that the second card will be an ace as well? Of the 51 remaining cards, 3 are aces. Therefore, p(B|A) = 3/51 = 1/17, and the probability of A and B is 1/13 1/17 = 1/221. 2. Probability of A or B If events A and B are mutually exclusive, then the probability of A or B is simply: p(A or B) = p(A) + p(B). What is the probability of rolling a die and getting either a 1 or a 6? Since it is impossible to get both a 1 and a 6, these two events are mutually exclusive. Therefore,
p(1 or 6) = p(1) + p(6) = 1/6 + 1/6 = 1/3
If the events A and B are not mutually exclusive, then
p(A or B) = p(A) + p(B) - p(A and B).
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(A and B) is subtracted. Example 11 What is the probability that a card selected from a deck will be either an ace or a spade? Solution The relevant probabilities are p(ace) = 4/52 p(spade) = 13/52 The only way 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.
