Question

A lot of 108 semiconductor chips contains 20 that are defective. Two are selected randomly, without replacement, from the lot. Round your answers to three decimal places (e.g. 98.765) a) What is the probability that the first one selected is defective? b) What is the probability that the second one selected is defective given that the first one was defective? c) What is the probability that both are defective? d) How does the answer to part (b) change (give the new value of the probability) if chips selected were replaced prior to the next selection?

Answer 1

Given , a lot of 108 semi-conductors has 20 that are defective. 2 are seected randomly without replacement.

a) The probability that the first one selected will be defective is 20/108 = 0.185

b.) Now we are given that the first item sampled was a defective.SInce we are working in a WITHOUT REPLACEMENT sampling scheme so the the chances of drawing a defective unit in succesive draws will decrease. Hence now there are only 19 defectives left.So the probability of second one is defective given that the first was defective is 19/107 = 0.176

c.) The probability that both will be defective will be found as the number of cases in which both drawn items were defective to the total number of possible ways of drawing 2 items from the 108 items.

                                          $=\binom{20}{2}/\binom{108}{2}$

                                           = 0.0329

d.) If the chips were to be replaced then the probability of drawing a defective in the second draw is independent of the fact whether a defective chip has turned up in the first draw or not.

                    The probability in part b would thus become 20/108 = 0.185