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Probability Question - Picking the cards

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Probability Question - Picking the cards [#permalink]

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New post 08 Aug 2009, 10:09
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I take this question from Princeton Review 2008 and not sure about the answer. Please suggest
Q:
Jerome wrote each of the integers 1 through 20, inclusive, on a separate index card. He placed the cards in a box, then drew cards one at a time randomly from the box, without returning the cards he had already drawn to the box. In order to ensure that the sum of all the cards he drew was even, how many cards did Jerome have to draw?

A. 19
B. 12
C. 11
D. 10
E. 3

From the choices, E can be first eliminated and I'm not quite sure what to do next. Please suggest.

thx!

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Re: Probability Question - Picking the cards [#permalink]

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New post 08 Aug 2009, 10:25
From the answer choices
A. 19 - 9 odd + 10 even. So sum is odd.
B. 12 - 3 odd + 9 even - Odd sum
C. 11 - 3 odd + 8 even - Odd sum.
D. 10 - 3 odd + 7 even - Odd sum.
E. 3 - 1 odd + 2 even - Odd sum.
So odd sum is possible in all of the answer choices.

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Re: Probability Question - Picking the cards [#permalink]

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New post 08 Aug 2009, 16:23
An odd + even results in an odd number.
So we must find how many numbers can make the sum till odd and we have to stop when the first number makes the sum even( such that no number remaining would make the sum odd).

Let the first number be even , so to make the sum odd we need an odd number (e + o = o). So the resulting sum is odd, so we need an even number to make the sum odd again, repeat this till we exhaust all the even numbers.

1st 2nd 3rd 4th .........

e o e e
sum=o sum= o sum =o

(e-even, o -odd)
when we exhaust all the even numbers , we would be having only odd numbers left so the first odd number would make the sum even. So we use all the even numbers and one odd number, since there are total 20 numbers we have 10 even numbers + 1 odd number = 11 and any first remaining odd number +1, gives an even sum , so the answer is 12. Hope this helps.

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Re: Probability Question - Picking the cards [#permalink]

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New post 08 Aug 2009, 19:58
But then drawing cards randomly, we cannot ensure that drawing 12 cards will always result in even sum. In this scenario how can we conclude?

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Re: Probability Question - Picking the cards [#permalink]

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New post 08 Aug 2009, 20:27
Ans is 19 cards...here is my logic

if you remove 19 cards we have 2 possibilities ie left over card is either even or odd...this means we can either have
(10e+9o....left cards is odd) or (10o+9e...left card is even)

case I : (10e+9o)
we know e+e = e also o+o=e
thus (e+e+e+e..10 times = even) also (o+o+o+...9 times =even)
thus both even + even = even

case II : 10o+9e.....with similar explination sum is even

thus we will always have even sum ..hence OA 19

**guys for rest of the confusion i m solving a option choice for 3 cards: for all other options we cannot conclude that sum is even ...( it can be both odd or even)

say we draw 3 cards : 4 possibilities
3e= even sum
3o=odd sum ( as o+o = e and e+o = odd)
1e+2o = even sum ( same logic as explained)
1o+2e = even sum ( same logic as explained)
Hence not satisfying condition
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Re: Probability Question - Picking the cards [#permalink]

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New post 08 Aug 2009, 22:16
Aleehsgonji wrote:
But then drawing cards randomly, we cannot ensure that drawing 12 cards will always result in even sum. In this scenario how can we conclude?


It is the worst case that we need 12 cards to get even sum.

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Re: Probability Question - Picking the cards [#permalink]

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New post 08 Aug 2009, 22:31
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Hey guys please check this link, it is the intrepretation of the problem that depends on the answer. And it is poorly worded too, so dont bother...

http://www.beatthegmat.com/hard-number- ... dex%20card

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Re: Probability Question - Picking the cards [#permalink]

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New post 09 Aug 2009, 07:19
Bhushan252,
O+O+O..9 times is not even but odd. So 19 is not the right choice.

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Re: Probability Question - Picking the cards [#permalink]

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New post 09 Aug 2009, 08:45
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Aleehsgonji wrote:
Bhushan252,
O+O+O..9 times is not even but odd. So 19 is not the right choice.


May be 20 is the correct answer, but not in choice..
check this link...
http://www.beatthegmat.com/hard-number- ... dex%20card

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Re: Probability Question - Picking the cards   [#permalink] 09 Aug 2009, 08:45
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