![]() ![]() Likewise, we can extend the addition rule for pairwise mutually exclusive events.Įxample 2: Counting Outcomes of Three Events Using the Addition Rule Then, we have 1 0 + 7 + 5 = 2 2 different options for lunch. We can see that there is no lunch item that is shared by any two restaurants, so the events of buying lunch from different restaurants are pairwise mutually exclusive. Now, a new restaurant, Sandwich Place, has opened, with 5 different sandwiches on its menu. Previously, we assumed that our town only has two restaurants: Pizza Shop, which offers 10 different menu items, and Soup Kitchen, which offers 7 different menu items. Let us return to our problem of buying a lunch. In other words, three or more events are pairwise mutually exclusive when there is no outcome that is shared by any two events. In this case, we need every pair of events to be mutually exclusive. To do this, we must first understand how the condition of mutual exclusivity can be extended when three or more events are involved. We can generalize the addition rule to apply it to more than two events. In the previous example, we applied the addition rule for two mutually exclusive events. Since we are choosing 2 girls from 6 girls, where the order of the girls does not matter, the number of different outcomes of this event is given by the combination □.Īpplying the addition rule, the number of ways to form a group that consists of 3 boys or 2 girls is Next, consider the event of forming a group consisting of 2 girls. Since we are choosing 3 boys from 10 boys, where the order of the boys does not matter, the number of different outcomes of this event is given by the combination □. We recall that the number of different ways to select □ objects from the total of □ distinct objects, where the order of the □ objects does not matter, is denoted by □.įirst, consider the event of forming a group consisting of 3 boys. We need to count the number of outcomes from both events. Hence, we can apply the addition rule to obtain our answer if we know the number of outcomes from both events. ![]() In this example, the events are “forming a group consisting of 3 boys” and “forming a group consisting of 2 girls.” We note that there cannot be a common outcome of the two events, which tells us that they are mutually exclusive. We recall the addition rule for two events: if two events are mutually exclusive, then the number of distinct outcomes from either of the two events is given by the sum of the numbers of distinct outcomes from the two events. What is the numerical expression that allows us to calculate how many ways there are of forming a group that consists of either 3 boys or 2 girls? ![]() ![]() Book traversal links for 3.Example 1: Counting Outcomes of Two Events Using the Addition Rule Incidentally, we'll see many more problems similar to this one here when we investigate the binomial distribution later in the course. I personally would not have wanted to solve this problem by having to enumerate and count each of the possible subsets. Where \(N(A)\) is the number of ways that he can get a 6 and a head, and \(N(\mathbf=1024\) possible subsets. The probability of his event \(A\), say, is: Therefore, he can use the classical approach of assigning probability to his event of interest. Similarly, because his coin is fair, he has an equally likely chance of getting a head or a tail. Because his die is fair, he has an equally likely chance of getting any of the numbers 1, 2, 3, 4, 5, or 6. When he rolls a fair six-sided die and tosses a fair coin. Roll Toss wants to calculate the probability that he will get: ![]()
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