Question 1135509
a is 102!/94!8! or 102C8=2.19 x 10^11 (2192950683 x 10^11) ANSWER a
No defective ones
92/102*91/101*90/100*89/99*88/98*87/97*86/96*85/95=0.4245 probability that none is defective so a 0.5755 probability at least one is defective.
1-(92C8/102C8)=1.2621418 x 10^11 ANSWER c, at least one defective

That means 9.308088718 x 10^10 ways there are no defective ones ANSWER b

exactly one is 10C1*92C7/102C8= 0.39948 probability or 8.76056 x 10^10 ways ANSWER d
Notice how the top part of the "choose" factor (hypergeometric distribution) adds to 102 and the bottom adds to 8.