A group of four components is known to contain two defectives. An inspector tests the components one at a time until the two defectives 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 .
Step1 of 2:
Let us consider an event ‘D’ be the defective components.
Let event ‘G’ no defective components.
We have 2 defectives out of 4 components.
We need to find the probability distribution for Y.
Step2 of 2:
Let Y denote the number of the test on which the second defective is found and it is given as:
P(both defective are found in the first two inspections) = P(Y = 2)
= P(Prob. Of first defective)P(Prob.Of
Hence, P(both defective are found in the first two inspections) = .
For Y = 3, we need to break it into two cases that is GDD and DGD.
P(Y = 3) = 2C1
Hence, P(Y = 3) = .
For Y = 3, we need to break it into three cases that is GGDD, DGGD, GDGD.
P(Y = 4) = 3C1
Hence, P(Y = 4) = .