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SUGGESTED ANSWER TO QUESTION 4-3 If three numbers are drawn, each of which can take a value from 0 to 9, there are 1000 possible drawings: 000, 001, 002, ..., 999. The numbers, 911, is one of the 1000 possible drawings. Thus, the probability that, on September 11 of any year, the numbers 911 will be drawn in the New York lottery is 1/1000. In fact, as the statistician John Allen Paulos (2002) noted, there are two drawings (morning and evening) each day in the New York lottery, so that the probability of drawing the numbers 911 is increased to just under 1/500. This can be calculated in the following way. The probability of drawing 911 in the morning is 1/1000, and the probability of drawing some other set of three numbers in the evening is 999/1000, which gives a probability of: 1/1000 * 999/1000 = 999/1,000,000 The probability of the reverse--of drawing 911 in the evening and some other set of three numbers in the morning--is also 999/1,000,000. Finally, the probability of drawing 911 in both the morning and evening is: 1/1000 * 1/1000 = 1/1,000,000 In order to get the probability that 911 will be drawn in at least one of the two drawings, we need to add the three separate probabilities: 999/1,000,000 + 999/1,000,000 + 1/1,000,000 = 1999/1,000,000 This number is one less than 2000/1,000,000; and 2000/1,000,000 reduces to 1/500. So the probabilty of selecting 911 in at least one of the two drawings on September 11th is just under 1/500. Although 1/500 may sound unlikely, we should expect that, eventually, the three numbers will be drawn in the New York lottery, as they were one year after the attacks. And Paulos (2002) argued that, given the fact that numbers are everywhere in our environments, the probability that the numbers 911 will arise somewhere on September 11th becomes quite high:
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