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Statistics

Notes Probability that a case is free of defective items

P (r = ) 0 = e  0.5 = 0.6065. . Hence the number of cases having no defective items = 0.6065 × 100 =
60.65

P
=
´
=
Similarly, (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 19: 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 (i) if less than 1% of the letters submitted by her
are rejected; (ii) 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].
Solution.
Let p be the probability of making a mistake in typing a word.
(i) Let the random variable r denote the number of mistakes per letter. Since 20 letters are
typed, r will follow Poisson distribution with mean = 20 × p.
Since less than 1% of the letters are rejected, it implies that the probability of making at
least one mistake is less than 0.01, i.e.,
P(r  1)  0.01 or 1 - P(r = 0)  0.01
Þ 1 – e -20p  0.01 or e -20p  0.99
Taking log of both sides

– 20p.log 2.72  log 0.99
 (20 0.4346 p´ )  1.9956
.
0 0044
– 8.692p  – 0.0044 or p  = 0 00051.
.
.
8 692
(ii) In this case r is a Poisson variate which denotes the number of mistakes per day. Since the
typist has to type 20 × 200 = 4000 words per day, the mean number of mistakes = 4000p.
It is given that there is no mistake on 90% of the days, i.e.,
P(r = 0) = 0.90 or e -4000p = 0.90
Taking log of both sides, we have

´
– 4000p log 2.72 = log 0.90 or  4000 0.4346p = 1.9542 =  0.0458
0.0458
 p = = 0.000026.
4000 0.4346
´

Example 20: A manufacturer of pins knows that on an average 5% of his product is
defective. He sells pins in boxes of 100 and guarantees that not more than 4 pins will be defective.
What is the probability that the box will meet the guaranteed quality?

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