Sanderson M. Smith

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An alert student picked up a mistake in a handout solution for a probability word problem.

Here is the problem as presented in the book. (The caps represent a clear statement that I overlooked when presenting my solution.)

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TRADING CARD PROBLEM:You collect trading cards, which have different scenes from a movie. For one movie, there are 90 different cards in the set, and you have all of them except the final scene. To try and get this card, you buy 10 packs of 8 cards each. ALL CARDS IN A PACK ARE DIFFERENT, and each of the cards is equally likely to be in a given pack. Find the probability that you will get the final scene.

SOLUTION(as presented by the student):For a given pack, the probability that it doesn't contain the desired card is

_{89}C_{8}/_{90}C_{8}= 0.9111111, or about 91%.If you buy 10 packs, the probability that none of them contain the desired card is (

_{89}C_{8}/_{90}C_{8})^{10}= 0.394197, or about 39.4%.Hence, the probability that at least one of the ten packs contains the card you want is 1 - (

_{89}C_{8}/_{90}C_{8})^{10}= 0.605803, or about 60.58%

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OK, here is basically the same problem, only the phrase "ALL CARDS IN A PACK ARE DIFFERENT" is removed. This creates a slightly different premise. This allows for the possibility that there could be duplicate cards in a pack of 8 cards. The solution I presented is below. (My solution is correct for the reworded problem, but not for the problem as presented in the text.)

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TRADING CARD PROBLEM:You collect trading cards, which have different scenes from a movie. For one movie, there are 90 different cards in the set, and you have all of them except the final scene. To try and get this card, you buy 10 packs of 8 cards each. And, each of the cards is equally likely to be in a given pack. Find the probability that you will get the final scene.

SOLUTION:This premise allows for the possibility that a pack might contain two or more of the same card. The problem is now equivalent to having 80 cards (8 cards in a pack, and you have 10 packs) all with an equally likely chance of being any one of the 90 scenes. OK, you have 80 cards, and the probability that none of them represent the card you want is (89/90)

^{80}= 0.409072, or about 40.91%. The probability that you get the card you want at least once is 1 - (89/90)^{80}= 0.5909, or about 59.09%.

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I offered kudos to the student who pointed out my (reading) error.

MATH POWER INCORPORATES THINKING, ANALYZING, INVESTIGATING ... AND GAINING MATH POWER INVOLVES LEARNING FROM YOUR MISTAKES. (OK, I learned I should have read the problem a bit more carefully.)

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