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102 example 1 in the license plate example what is

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102
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EXAMPLE : (1) In the license plate example, what is the probability of getting a licence plate with all letters the same and all digits the same? (2) What is the probability of having a license plate with all letters the same or all digits the same? 103
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SOLUTION: (1) We already know how many different license plates are possible: 17,576,000. There are only 26 ways of getting the same three letters: AAA, BBB, up through ZZZ. There are only 10 ways of getting the same three digits (for example, 444). The number of plates having the same three letters and the same three digits is 26 10 260 such plates. The probability of having such a plate is 260/17,576,000 .00001479. 104
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(2) As noted just noted, there are 26 possibilities for getting all three letters the same. Each of these can be combined with the 10 3 1,000 possible digit combinations. So there are 26,000 licence plates with all three letters the same. Similarly, there are 26 3 10 175,760 license plates with the same three digits (000, 111, and so on). But we have double counted the number of plates from part (1), so we subtract these off. The resulting probability is 26,000 175,760 260 17,576,000 .0115 105
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EXAMPLE : Suppose an urn contains 10 balls, 6 black and 4 red. If 5 balls are selected without replacement from the urn, what is the probability of getting 3 black and 2 red balls? 106
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SOLUTION: The number of ways we can choose 5 balls from 10 (without regard to order) is 10 5  . The number of ways we can choose 3 black balls from 6 is 6 3 and the number of ways we can choose 2 red balls from 4 is 4 2 . Therefore, the probability of getting 3 black and 2 red balls is 6 3 4 2 10 5 4 6 252 2 21 In the previous example, we get the same answer if we approach the problem from an ordered perspective. 107
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EXAMPLE : The “birthday problem.” Assuming that each of the 365 days (ignoring leap years) is equally likely for a birthday, what is the probability that in a group of k people at least two have the same birthday? Let A be the event that at least two of the k people have the same birthday. Then A c is the event that all k people have different birthdays. 108
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P A c is actually easier to compute than P A . Why? Given k people, the number of different birthday sequences is 365 365 365 365 k because each person can be born on any of the 365 days. The number of elements in A c is 365 P k 365 364  365 k 1 365!/ 365 k ! 109
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Therefore, P A c P all k people have different birthdays 365 364  365 k 1 365 k and P A 1 365 364  365 k 1 365 k When k 10, P A .117; when k 20, P A .411; when k 23, P A .507; when k 30, P A .706; when k 40, P A .891; and when k 50, P A .970. 110
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The previous probabilities are not the probabilities that someone shares, say, your particular birthday. For person k , the probability that at least one of the k 1 remaining people shares the same birthday is 1 P none of the k 1 people shares k ’s birthday 1 P person 1 has different b’day  P person k 1 has different b’day 1 364 365 k 1 111
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We get this by letting, say, B j
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102 EXAMPLE 1 In the license plate example what is the...

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