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Unit 14: Poisson Probability Distribution
Solution: Notes
10
The random variable denotes the number accidents per day. Thus, we have m 0.2 . The
50
required probability is given by
2
0.2
0.2
P r 3 1 P r 2 1 e 1 0.2 1 0.8187 1.22 0.00119.
2!
Example: A car hire firm has two cars which it hire out every day. The number of
demands for a car on each day is distributed as a Poisson variate with mean 1.5. Calculate the
proportion of days on which neither car is used and the proportion of days on which some
-1.5
demand is refused. [ e = 0.2231]
Solution:
When both car are not used, r = 0
1.5
P r 0 e 0.2231. Hence the proportion of days on which neither car is used is 22.31%.
Further, some demand is refused when more than 2 cars are demanded, i.e., r > 2
2 1.5 r 2
e 1.5 1.5
P r 2 1 P r 2 1 1 0.2231 1 1.5 0.1913.
r! 2!
r 0
Hence the proportion of days is 19.13%.
Example: A firm produces articles of which 0.1 percent are usually defective. It packs
them in cases each containing 500 articles. If a wholesaler purchases 100 such cases, how many
cases are expected to be free of defective items and how many are expected to contain one
defective item?
Solution:
The Poisson variate is number of defective items with mean
1
m 500 0.5.
1000
Probability that a case is free of defective items
0.5
P r 0 e 0.6065.Hence the number of cases having no defective items = 0.6065 × 100
= 60.65
Similarly, P r 1 e 0.5 0.5 0.6065 0.5 0.3033.Hence the number of cases having one
defective item are 30.33.
Example: A manager accepts the work submitted by his typist only when there is no
mistake in the work. The typist has to type on an average 20 letters per day of about 200 words
each. Find the chance of her making a mistake (1) if less than 1% of the letters submitted by her
are rejected; (2) if on 90% of days all the work submitted by her is accepted. [As the probability
of making a mistake is small, you may use Poisson distribution. Take e = 2.72].
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