(a) The probability of the first cartridge being defective is 6/85. There are 84 remaining, and 5 are defective, so the probability of selecting another defective one is 5/84; that leaves 83, 4 defective. The probability of the third being defective is 4/83. The combined probability is 6/85*5/84*4/83=0.00020 approx. (approx 0.02%).
(b) The probability of the first cartridge being OK is 79/85. Now there are 84, including 6 defects, leaving 78. The probability of the second being OK is 78/84 and of the third being OK it's 77/83. The probability of all 3 being OK is therefore 79/85*78/84*77/83=0.800638 approx. Therefore, the probability of at least one being defective is 1-0.800638=0.19936 approx (19.936%), because all other possibilities include at least one defect.