Question

Suppose that two defective refrigerators have been included in a shipment of eight refrigerators. The buyer begins to test the eight refrigerators one at a time. Let the random variable Y represent the number of defective refrigerators found after three refrigerators have been tested. Compute the probability distribution of Y.

I did a similar problem to this but worked out a sample space etc. and solved for certain probabilities however what confused me then and what confuses me now is why is the sample space not 2 ^ (8) and rather 8 C 2

Answer 1

It seems that you are trying to apply the logic that since each refrigerator can either be okay or defective (2 options) and there are 8 of them, the total number of possibilities are 2 ^ (8). But as you should be able to observe from the preceding line, it is the set of combinations of all the possibilities (no defective refrigerator to all of them defective). But we have been specifically told that only two of the eight are defective. Thus, the sample space should only consist of the number of elements equal to the number of ways in which 2 refrigerators can be chosen out of 8, i.e. 8C2. Let's try to understand it in a simpler way.

Let D represent defective and O represent okay refrigerators. Then our sample space would be the set of all possible combinations of 2 Ds and 6 Os as given in the problem statement.

1_2 _3 _4 _5 _6 _7 _8 _

The above 8 blank spaces can contain the 2 Ds and 6 Os. The 2 Ds can be placed in the 8 blank spaces in 8 C 2 ways. After that, the 6 Os can be placed on the remaining spaces as they are all the same. Thus 8 C 2 is the number of elements in the sample space.

Now, Y is the number of defective refrigerators found after three refrigerators have been tested. Since the maximum value of Y can be 2, Y can take any of the values from {0,1,2}

Y = 0 implies that the combinations are of the form O O O _ _ _ _ _ _, where the five blank spaces must contain the 2 Ds. The number of ways in which it can happen is 5 C 2.

Y = 1 implies that in the first three places, there is 1 D and 2 Os and in the remaining 5 places, there is 1 D and 4 Os. The number of ways in which it can happen is (3 C 1 ) * (5 C 1).

Y = 2 implies that both the 2 Ds are in the first 3 places and the rest are Os. This can happen in 3 C 2 ways.