SECTION 13.3 Probability 931 Number of People 5 101520212223242530405060 70 80 90 Probability That Two or More Have the Same Birthday 0.027 0.117 0.253 0.411 0.444 0.476 0.507 0.538 0.569 0.706 0.891 0.970 0.994 0.99916 0.99991 0.99999 Birthday Problem What is the probability that in a group of 10 people, at least 2 people have the same birthday? Assume that there are 365 days in a year and that a person is as likely to be born on one day as another, so all the outcomes are equally likely. Solution EXAMPLE 10 First determine the number of outcomes in the sample space S. There are 365 possibilities for each person’s birthday. Since there are 10 people in the group, there are 36510 possibilities for the birthdays. [For one person in the group, there are 365 days on which their birthday can fall; for two people, there are ⋅ = 365 365 3652 pairs of days; and, in general, using the Multiplication Principle, for n people there are 365n possibilities.] So ( ) = n S 36510 We wish to find the probability of the event E: “at least two people have the same birthday.” It is difficult to count the elements in this set; it is much easier to count the elements of the complementary event E: “no two people have the same birthday.” Find ( ) n E as follows: Choose one person at random. There are 365 possibilities for their birthday. Choose a second person.There are 364 possibilities for this birthday, if no two people are to have the same birthday. Choose a third person. There are 363 possibilities left for this birthday. Finally, arrive at the tenth person. There are 356 possibilities left for this birthday. By the Multiplication Principle, the total number of possibilities is … ( ) = ⋅ ⋅ ⋅ ⋅ n E 365 364 363 356 The probability of the event E is P E n E n S 365 364 363 . . . 356 365 0.883 10 ( ) ( ) ( ) = = ⋅ ⋅ ⋅ ⋅ ≈ The probability of two or more people in a group of 10 people having the same birthday is then ( ) ( ) = − ≈ − = P E P E 1 1 0.883 0.117 The birthday problem can be solved for any group size.The following table gives the probabilities for two or more people having the same birthday for various group sizes. Notice that the probability is greater than 1 2 for any group of 23 or more people. Now Work PROBLEM 71 Historical Feature Set theory, counting, and probability first took form as a systematic theory in an exchange of letters (1654) between Pierre de Fermat (1601–1665) and Blaise Pascal (1623–1662). They discussed the problem of how to divide the stakes in a game that is interrupted before completion, knowing how many points each player needs to win. Fermat solved the problem by listing all possibilities and counting the favorable ones, whereas Pascal made use of the triangle that now bears his name. As mentioned in the text, the entries in Pascal’s triangle are equivalent to ( ) C n r , . This recognition of the role of ( ) C n r , in counting is the foundation of all further developments. The first book on probability, the work of Christiaan Huygens (1629–1695), appeared in 1657. In it, the notion of mathematical expectation is explored. This allows the calculation of the profit or loss that a gambler might expect, knowing the probabilities involved in the game (see the Historical Problem that follows). (continued) Blaise Pascal (1623–1662) Credit: GeorgiosArt/ iStock/Getty Images
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