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A dozen computer diskettes, including 2 that are defective, are returned to a retailer. A random sample of 3 is selected and tested. Find the expected number of defective diskettes in a sample.

Answer:

Let X be the random variable that is the number of defective diskettes in any sample. We need to find the probability distribution of X. The number of ways of obtaining 3 items from 12 is C12,3 order is not important. This is the number of equally likely events in the sampling. A sample will have either 0, 1, or 2 defective diskettes. These are the possible values of X.

To obtain a sample with 0 defective diskettes we have to remove the 2 defective diskettes from consideration, so we have the total number of possible combinations for 10 diskettes taken 3 at a time.

In this case to obtain a sample with only 1 defective diskette we need to find the number of possible ways we can get 2 nondefective diskettes in a sample which is the 10 nondefective diskettes taken 2 at a time, then how many ways could we draw one of the 2 defective diskettes one at a time. So the total number of ways to get a sample of 2 nondefective diskettes and 1 defective is the C10,2 times C2,1. This is similar to the multiplication rule of probability.

A similar process is done for the number of ways finding 2 defective diskettes in a sample.

The probability distribution of X is

x

1

1

2

p(x)

12/22

9/22

1/22

 

 

The expected number of defective diskettes is

E(X) = 0(12/22) + 1(9/22) + 2(1/22) = 0 + 9/22 + 2/22 = 11/22 = 1/2

There would be 0.5 defective diskettes in each sample.