Each of 20 identical cards is numbered with exactly one of the numbers 1,2,3,.....20. One card is drawn randomly and it is known that the number on the card is less than 13. What is the probability that the number on the card is an even number?
Probability On Numbers
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0That's just as if you had started with numbers $1, 2, \ldots, 12$ righ taway. – 2012-09-18
1 Answers
If the card's value is less than 13, then given the other information its value must be in the range 1..12. This range has an even number of elements (12), and thus exactly half of them (6) will be even, so the probability of the card having an even value is $\dfrac{6}{12} = \dfrac{1}{2} =.5$.
The facts that (1) the set actually contains more than 12 cards, and that (2) the overall probability of drawing any card in the full set with an even number is also .5, are immaterial; you are told the card's value is in a subset of the full set (cards < 13), thus you are determining the probability of it being in a subset of the subset (even cards < 13), as if the subset were the full set of cards. If we did not know the card was < 13, and were asked what the probability was of the card being an even number < 13, that's a very different question.
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0What is assumed but not shown in the answers is that given X<13 the numbers 1-12 are all equally likely. This seems intuitive but is best proven mathematically. It also seems intuitive to many that the probablity of winning by switching in the Monty Hall problem is 1/2 (in that case reasoning incorrectly that the conditioning leaves the doors equally likely to have the prize). – 2012-09-18