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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?

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That's just as if you had started with numbers $1, 2, \ldots, 12$ righ taway. –  Hagen von Eitzen Sep 18 '12 at 17:28

1 Answer 1

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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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but i thought the total sample is 20. I mean n(s)=20 right? But why you divided by 12 instead of 20?? –  David Sep 18 '12 at 17:36
@David: When you are told that the number is less than 13, the sample is reduced. It is not possible that you drew 15, for example. That is the point of KeithS's argument. –  Ross Millikan Sep 18 '12 at 17:45
ok i got. Thanks btw guys –  David Sep 18 '12 at 17:50
For this problem, the approach above is clear and quick. One can also complicate things. Let $L$ be the event "less than $13$" and $E$ the event "even". We want $\Pr(E|L)$ (probability of $E$ given $L$). Then one can use the basic formula $\Pr(X|Y)=\frac{\Pr(X\cap Y)}{\Pr(Y)}$. For more complicated problems, this "conditional probability" approach is very useful, but it would be gross overkill here. –  André Nicolas Sep 18 '12 at 18:04
yeah thanks andre for this approach. I feel this approach able to make me understand better. But how am I supposed to know either P(E|L) or P(L|E)?? How to determine it? –  David Sep 18 '12 at 18:10

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