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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 “occurring in
threes” 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 “normal” or “extremely short.” There are
twenty-nine in the first 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 qualifies as a rare event:
P(Two extremely short separations occur consecutively at least once)
30 6 30 5 30 4
· 1 + 5 · 2 + 4 · 3
= 0.61
= 6
So, despite what our intuitions might tell us, the phenomenon of bad things coming
in threes is neither mysterious nor uncommon or unexpected.
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 five 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 five minutes or more to see another
% ∞
0.67e −0.67y dy
P(Y > 5) =
e−u du
(where u = 0.67y)
= −e−u '3.35 = e−3.35
= 0.035
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?