Agroup 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.

when Y = 2, P(2) = 1/6 = 0.167

when Y = 3, P(3) = 2/6 = 0.333

when Y = 4, P(4) = 3/6 = 0.5

Step-by-step explanation:

Total number of possible ways to choose 2 components for defectives out of 4 components = 4C2 = 6

Y can only take values of 2, 3 and 4

So, Probability of Y = P(Y) = 1C(Y-1) / 6

when Y = 2, P(2) = 1/6 = 0.167

when Y = 3, P(3) = 2/6 = 0.333

when Y = 4, P(4) = 3/6 = 0.5

step-by-step explanation: