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prob.part41.85_86

# prob.part41.85_86 - 3.1 PERMUTATIONS meat.5 soup.8...

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3.1. PERMUTATIONS 77 (start) soup meat fish vegetable juice .8 .2 .2 .3 .3 .4 .5 .3 meat fish vegetable ω ( ω ) ω ω ω ω ω ω .4 .24 .16 .06 .08 .06 m 1 2 3 4 5 6 Figure 3.2: Two-stage probability assignment. that a customer chooses meat is m ( ω 1 ) + m ( ω 4 ) = . 46. We shall say more about these tree measures when we discuss the concept of conditional probability in Chapter 4. We return now to more counting problems. Example 3.2 We can show that there are at least two people in Columbus, Ohio, who have the same three initials. Assuming that each person has three initials, there are 26 possibilities for a person’s first initial, 26 for the second, and 26 for the third. Therefore, there are 26 3 = 17 , 576 possible sets of initials. This number is smaller than the number of people living in Columbus, Ohio; hence, there must be at least two people with the same three initials. We consider next the celebrated birthday problem—often used to show that naive intuition cannot always be trusted in probability.

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