Introduction to Probability and Statistics for Engineers and Scientists

5.2The Poisson Random Variable 151

6. The number ofα-particles discharged in a fixed period of time from some radioactive
   particle.

Each of the foregoing, and numerous other random variables, is approximately Poisson
for the same reason — namely, because of the Poisson approximation to the binomial. For
instance, we can suppose that there is a small probabilitypthat each letter typed on a page
will be misprinted, and so the number of misprints on a given page will be approximately
Poisson with meanλ=npwherenis the (presumably) large number of letters on that
page. Similarly, we can suppose that each person in a given community, independently, has
a small probabilitypof reaching the age 100, and so the number of people that do will have
approximately a Poisson distribution with meannpwherenis the large number of people
in the community. We leave it for the reader to reason out why the remaining random
variables in examples 3 through 6 should have approximately a Poisson distribution.

EXAMPLE 5.2a Suppose that the average number of accidents occurring weekly on a par-
ticular stretch of a highway equals 3. Calculate the probability that there is at least one
accident this week.

SOLUTION LetXdenote the number of accidents occurring on the stretch of highway in
question during this week. Because it is reasonable to suppose that there are a large number
of cars passing along that stretch, each having a small probability of being involved in
an accident, the number of such accidents should be approximately Poisson distributed.
Hence,

P{X≥ 1 }= 1 −P{X= 0 }

= 1 −e−^3

30

0!

= 1 −e−^3

≈.9502 ■

EXAMPLE 5.2b Suppose the probability that an item produced by a certain machine will
be defective is .1. Find the probability that a sample of 10 items will contain at most one
defective item. Assume that the quality of successive items is independent.

SOLUTION The desired probability is

( 10

0

)

(.1)^0 (.9)^10 +

( 10

1

)
(.1)^1 (.9)^9 =.7361, whereas
the Poisson approximation yields the value

e−^1

10

0!

+e−^1

11

1!

= 2 e−^1 ≈.7358 ■

EXAMPLE 5.2c Consider an experiment that consists of counting the number ofαparti-
cles given off in a one-second interval by one gram of radioactive material. If we know