Midterm 2x2 crosstab solutions

a) Here is the crosstab:

	Supplier A	Supplier B	Total
Defective	3	8	11
Not defective	17	72	89
Total	20	80	100

Since P(defective | A) = 15%, then P(A and defective)/P(A) = 0.15 from which we get P(A and defective)/0.2 = 0.15 and P(A and defective) = (0.2)(0.15) = 0.03 = 3%. Similarly, P(B and defective) = (0.8)(0.1) = 0.08 = 8%

b) P(A and not defective) = 17% from crosstab.

c) P(B | not defective) = 72/89 = 80.9%

d) P(B or not defective) = P(B) + P(not defective) – P(B and not defective)

= 80% + 89% - 72% = 97%

e) P(not defective | A) = 17/20 = 85%. Since we are told at the beginning of the problem that 15% of the parts from A are defective, it stands to reason that the remaining 85% would not be defective.

f) P(not A and not defective) = P(B and not defective) = 72% from crosstab.

g) 11% are defective; 3% are from A and 8% are from B. The answer is supplier B.

h) P(A and defective) = 3% from the crosstab. Therefore the remaining percentage is 97%.