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This is an exercise from a probability textbook: A frazzle is equally likely to contain $0,1,2,3$ defects. No frazzle has more than three defects. The cash price of each frazzle is set at \$ $10-K^2$, where $K$ is the number of defects in it. Gummed labels, each representing $\$ 1$, are placed on each frazzle to indicate its price. What is the probability that a randomy selected label will end up on a frazzle which has exactly two defects?


Since the frazzles are equally likely to have $0,1,2,3$ defects, I may argue that a label is equally likely to appear on any of them. On the other hand, frazzles with less defects are more expensive, therefore requiring more labels, from this perspective, a label is most likely to appear on a frazzle with no defects. I am confused here.

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It is not equally likely to go on any of the frazzles, because more labels will go to the frazzles with 0 defects than those with 3 defects, for example.

0,1,2,3 defects draws 10, 9, 6 and 1 labels respectively. So say you had 4 million frazzles. Since 0,1,2 or 3 defects are equally likely, suppose you have 1 million of each type. Then you have 10 million labels on those with 0 defects, 9 million labels on those with 1 defect, 6 million labels on those with 2 defects, and 1 million labels on those with 3 defects.

So you used a total of 26 million labels, and 6 million of those labels went to frazzles with exactly two defects. Thinking about this example should lead you to understanding what the answer to your question is.

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Got it, thank you very much! –  qed Dec 16 '12 at 15:15

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