Page 232 - DMGT524_TOTAL_QUALITY_MANAGEMENT

Unit 14: Statistical Process Control

The set of all values that a discrete random variable can assume with nonzero probability is Notes
either finite or countably infinite because the sum of uncountably many positive real numbers
(which is the smallest upper bound of the set of all finite partial sums) always diverges to
infinity. Typically, the set of possible values is topologically discrete in the sense that all its
points are isolated points. But, there are discrete random variables for which this countable set is
dense on the real line.

Combination Formula and its Application to Probability Calculations

Suppose we have a lot of 50 articles which contain 3 defective articles. If we draw a sample of
5 articles from the lot the respective probabilities of getting 0, 1, 2 and three defective articles in
a sample of 5 can be calculated as under. A sample containing 0 defective articles will come only
from 47 good articles. In this case no article will come from 3 defective articles. Number of ways
in which this can happen can be calculated as under:

47! (47)(46)(45)(44)(43)
47 C = = = 1533939
5
5!42! (5) (4) (3) (2) (1)
A sample containing 1 defective article will have 4 articles coming from 47 good articles and
1 defective article coming from 3 defective articles. Number of ways can be calculated as under:

47!3! (47)(46)(45)(44)
47 C C = = = 535095
3
4 1
(4!43!)(2!1!) (4)(2)(1)
Similarly the number of ways in which 2 and 3 defective articles can be drawn can be calculated
which come out to be 48645 and 1081 respectively. Further, the number of ways in which
5 articles can be drawn from 50 articles can be calculates as under:
(50)(49)(48)(47)(46)
50 C = = 2118760
5
(5)(4)(3)(2)(1)
Probabilities of 0, 1, 2 and 3 defectives can be calculated now as under.

1533939
P = = 0.72398
0
2118760
535095
P = = 0.25255
1
2118760
48645
P = = 0.02296
2
2118760
1081
P = = 0.00051
3
2118760
Total = 1.00000
Hypergeometric Probability Distribution
We have calculated the probability of getting r = 1, 2 or 3 defectives by drawing a sample of size
n = 5 from a lot N = 50 which contained D = 3 defective items. Hypergeometric probability law
describes such problems as under:

DN D C
−
C
P(r/N/D/n) = R n r −
N C N

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