By Marcel B. Finan

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**Extra resources for A Probability Course for the Actuaries: A Preparation for Exam P 1**

**Example text**

Solution. There are (6 - 1)! = 5! = 120 ways to seat 6 persons at a circular dinner table. Combinations In a permutation the order of the set of objects or people is taken into account. However, there are many problems in which we want to know the number of ways in which k objects can be selected from n distinct objects in arbitrary order. For example, when selecting a two-person committee from a club of 10 members the order in the committee is irrelevant. That is choosing Mr. A and Ms. B in a committee is the same as choosing Ms.

Our sample space consists of those students who did not get the flu shot. Let E be the set of those students without the flu shot who did get the flu. 45. 55 The union of two events A and B is the event A ∪ B whose outcomes are either in A or in B. The intersection of two events A and B is the event A ∩ B whose outcomes are outcomes of both events A and B. Two events A and B are said to be mutually exclusive if they have no outcomes in common. In this case A ∩ B = ∅ and P (A ∩ B) = P (∅) = 0. 2 Consider the sample space of rolling a die.

2 There are 6! ” In how many of them is r the second letter? Solution. Let r be the second letter. Then there are 5 ways to fill the first spot, 4 ways to fill the third, 3 to fill the fourth, and so on. There are 5! 3 Five different books are on a shelf. In how many different ways could you arrange them? Solution. The five books can be arranged in 5 · 4 · 3 · 2 · 1 = 5! = 120 ways Counting Permutations We next consider the permutations of a set of objects taken from a larger set. Suppose we have n items.