A quality control engineer tests the quality of produced com

A quality control engineer tests the quality of produced computers. Suppose that 5% of computers have defects, and defects occur independently of each other.

a. What is the expected number of defective computers in a shipment of twenty?

b. Find the probability of exactly 3 defective computers in a shipment of twenty

c. Find the probability that the engineer has to test at least 5 computers in order to find a defective one.

Solution

a)

E(x) = n p = 20*0.05 = 1 [answer]

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b)

Note that the probability of x successes out of n trials is          
          
P(n, x) = nCx p^x (1 - p)^(n - x)          
          
where          
          
n = number of trials =    20      
p = the probability of a success =    0.05      
x = the number of successes =    3      
          
Thus, the probability is          
          
P (    3   ) =    0.059582148 [answer]

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c)

This is a cumulative negative binomial distribution.

If we define \"success\" as getting a defective unit, then using technology, we can get

P(at least 4 fails before 1 success) = 0.81450625 [answer]

[If you use Excel, you can use =NEGBINOM.DIST(3 (maximum fails), 1 (success), 0.05, TRUE) to get at most 3 fails before 1 success.

Then, get its complement (at least 4 fails) to get the final answer. We need at least 4 fails so that we have tested 5 when we found a success (the success being the 5th.)]