In: Statistics and Probability
A worn machine is known to produce 10% defective components. If the random variable X is the number of defective components produced in a run pf 3 components, find the probabilities that X takes the values 0 to 3.
Suppose now that a similar machine which is known to produce 1% defective components is used for a production run of 40 components.We wish to calculate the probability that two defective items are produced. Essentially we are assuming thatX~B(40,0.01) and we use both the binomial distribution and its Poisson approxiamation for comparison.
Solution
Assuming that the production of components is independent and that the probability p = 0.1 of producing a defective component remains constant, the following table summarizes the production run. We let G represent a good component and let D represent a defective component. Note that since we are only dealing with two possible outcomes, we can say that the probability q of the machine producing a good component is 1 − 0.1 = 0.9. More generally, we know that q+p = 1 if we are dealing with a binomial distribution.
From this table it is easy to see that
P(X = 0) = (0.9)3
P(X = 1) = 3 × (0.9)2 (0.1)
P(X = 2) = 3 × (0.9)(0.1)2
P(X = 3) = (0.1)3
Clearly, a pattern is developing. In fact you may have already realized that the probabilities we have found are just the terms of the expansion of the expression (0.9 + 0.1)3
since (0.9 + 0.1)3 = (0.9)3 + 3 × (0.9)2 (0.1) + 3 × (0.9)(0.1)2 + (0.1)3
Using the binomial distribution we have the solution
P(X = 2) = 40C2(0.99)40−2 (0.01)2 =(40*39)/(1*2)× 0.9938 × 0.012 = 0.0532
Note that the arithmetic involved is unwieldy. Using the Poisson approximation we have the solution
P(X = 2) = e −0.4 (0.42/2!)= 0.0536
Note that the arithmetic involved is simpler and the approximation is reasonable