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By Marcel B. Finan

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

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Indistinguishable) objects can be distributed into k boxes is n+k−1 k−1 = n+k−1 n . Proof. Imagine the n identical objects as n stars. Draw k−1 vertical bars somewhere among these n stars. This can represent a unique assignment of the balls to the boxes. Hence, there is a correspondence between a star/bar picture and assignments of balls to boxes. So how many ways can you arrange n identical dots and k − 1 vertical bar? The answer is given by n+k−1 k−1 = n+k−1 n It follows that there are C(n + k − 1, k − 1) ways of placing n identical objects into k distinct boxes.

Each digit may be either 1,2 or 3. Use a tree diagram to show all the possible outcomes and tell how many different numbers can be selected. Solution. 31 32 COUNTING AND COMBINATORICS The different numbers are {11, 12, 13, 21, 22, 23, 31, 32, 33} Of course, trees are manageable as long as the number of outcomes is not large. If there are many stages to an experiment and several possibilities at each stage, the tree diagram associated with the experiment would become too large to be manageable. For such problems the counting of the outcomes is simplified by means of algebraic formulas.

Solution. 6 What is the probability of rolling a 3 or a 4 with a fair die? Solution. 7 In a room containing n people, calculate the chance that at least two of them have the same birthday. Solution. We have P(Two or more have birthday match) = 1 - P(no birthday match) Since each person was born on one of the 365 days in the year, there are (365)n possible outcomes (assuming no one was born in Feb 29). 5 It is important to keep in mind that the above definition of probability applies only to a sample space that has equally likely outcomes.

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