A purchaser of electrical components buys several lots, where each lot comes in a bundle of 6 (distinct) components. It is purchaser's policy to randomly select a lot and inspect 3 of the 6 components (also randomly selected). The entire shipment of lots is accepted only if all 3 are nondefective in the selected lot. The supplier knows that the number of defectives k in each lot is approximately distributed as: P[k ] = 0.36/k!

(a) (i) In how many ways can the 3 inspected components be selected ?

(ii) In how many ways can the shipment be accepted if the lot to be inspected contains k defective components ?

(b) Hence show that P[ accept lot ] = 0.579

(c) From your work in part (b) it follows that the supplier can increase his chances of getting the shipment accepted by making it more likely (eg., increase P[k ] = 0.50/k! ) that a lot contains k defective components !? Explain this apparent contradiction.