A group of four components is known to contain two defectives. An inspector tests the components one at a time until the two def
ectives are located. Once she locates the two defectives, she stops testing, but the second defective is tested to ensure accuracy. Let Y denote the number of the test on which the second defective is found. Find the probability distribution for Y.
We will assume A×B = B×A and show that A and B are necessarily the same.
Assume A×B = B×A. Let a ∈ A and b ∈ B. Then (a,b) ∈ A×B. Since A×B = B×A, we have (a,b) ∈ B×A. That is, a ∈ B and b ∈ A. Therefore, a ∈ A implies a ∈ B and b ∈ B implies b ∈ A. So A = B, as we had set out to do.
