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238 Chapter 4 Special Distributions they should come one month apart. But that is simply not the way random events behave, as Theorem 4.2.3 clearly shows. Look at the entries in Table 4.2.5. The average of those thirty-six (randomly occurring) eruption separations was 37.7 months, yet seven of the separations were extremely short (less than or equal to six months). If two of those extremely short separations happened to occur consecutively, it would be tempting (but wrong) to conclude that the eruptions (since they came so close together) were âoccurring in threesâ for some supernatural reason. Using the combinatorial techniques discussed in Section 2.6, we can calculate the probability that two extremely short intervals would occur consecutively. Think of the thirty-six intervals as being either ânormalâ or âextremely short.â There are twenty-nine in the ï¬rst group and seven in the second. Using the method described in Example 2.6.21, the probability that two extremely short separations would occur consecutively at least once is 61%, which hardly qualiï¬es as a rare event: P(Two extremely short separations occur consecutively at least once) 30 6 30 5 30 4 Â· 1 + 5 Â· 2 + 4 Â· 3 = 0.61 = 6 36 29 So, despite what our intuitions might tell us, the phenomenon of bad things coming in threes is neither mysterious nor uncommon or unexpected. Example 4.2.5 Among the most famous of all meteor showers are the Perseids, which occur each year in early August. In some areas the frequency of visible Perseids can be as high as forty per hour. Given that such sightings are Poisson events, calculate the probability that an observer who has just seen a meteor will have to wait at least ï¬ve minutes before seeing another one. Let the random variable Y denote the interval (in minutes) between consecutive sightings. Expressed in the units of Y , the forty-per-hour rate of visible Perseids becomes 0.67 per minute. A straightforward integration, then, shows that the probability is 0.035 that an observer will have to wait ï¬ve minutes or more to see another meteor: % â 0.67e â0.67y dy P(Y > 5) = % = 5 â eâu du (where u = 0.67y) 3.35 'â = âeâu '3.35 = eâ3.35 = 0.035 Questions 4.2.26. Suppose that commercial airplane crashes in a certain country occur at the rate of 2.5 per year. (a) Is it reasonable to assume that such crashes are Poisson events? Explain. (b) What is the probability that four or more crashes will occur next year? (c) What is the probability that the next two crashes will occur within three months of one another? 4.2.27. Records show that deaths occur at the rate of 0.1 per day among patients residing in a large nursing home. If someone dies today, what are the chances that a week or more will elapse before another death occurs?