In a semiconductor manufacturing process, three wafers from a lot are tested. Each wafer is classified as pass or fail. Assume that the probability that a wafer passes the test is 0.6 and that wafers are independent. Determine the probability mass function of the number of wafers from a lot that pass the test.

P(X = 0) = 0.064

P(X = 1) = 0.288

P(X = 2) = 0.432

P(X = 3) = 0.216

Step-by-step explanation:

Given that the probability of a wafer passing the test = 0.6, hence the probability of a wafer failing the test = 1 - 0.6 = 0.4.

Let X be the number of wafers that pass the test, since they are 3 wafers, hence X = 0, 1, 2, 3

Probability that all the three wafers failed the test = P(X = 0) = 0.4 × 0.4 × 0.4 = 0.064

Probability that all one wafer passed the test = P(X = 1) = (0.6 × 0.4 × 0.4) + (0.4 × 0.6 × 0.4) + (0.4 × 0.4 × 0.6) = 0.288

Probability that two wafers passed the test = P(X = 2) = (0.6 × 0.6 × 0.4) + (0.4 × 0.6 × 0.6) + (0.6 × 0.4 × 0.6) = 0.432

Probability that all three wafers passed the test = P(X = 3) = (0.6 × 0.6 × 0.6) = 0.216

hi,

i'm 75% sure the answer is 1,1 to both lines.

have a great day