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# A Bin contains card numbered from 1 to 20. If two cards are

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A Bin contains card numbered from 1 to 20. If two cards are [#permalink]

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03 Jun 2005, 14:43
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A Bin contains card numbered from 1 to 20. If two cards are drawn what is the probability the sum of the numbers on the cards is even ?

1/3
7/18
4/9
9/19
1/2

I think the answer should be 1/2 as opposed to 9/19 if am not wrong. For numbers 1-20, the sum of any 2 numbers is either even or odd, hence asked probability is 1/2

Can anyone point the flaw in the reasoning
thanks
Director
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Re: Probability Q in Challenge 17 [#permalink]

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03 Jun 2005, 15:54
bmw wrote:
A Bin contains card numbered from 1 to 20. If two cards are drawn what is the probability the sum of the numbers on the cards is even ?

1/3
7/18
4/9
9/19
1/2

I think the answer should be 1/2 as opposed to 9/19 if am not wrong. For numbers 1-20, the sum of any 2 numbers is either even or odd, hence asked probability is 1/2

Can anyone point the flaw in the reasoning
thanks

It should be 9/19.

There are 10 even and 10 odd numbered cards

So to have the sum even either both of them should be even or both odd.

Even cases are same as odd cases

Lets take EVEN

there are 10 even numbers and you draw 2 out of them.

So first card is drawn in 10/20 ways and the second card in 9/19 ways

So the prob of drawing even card is 10*9/20*19

Same is the case with odd numbered cards

So total = 2*(10*9)/(20*19 ) = 9/19
Manager
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03 Jun 2005, 16:20
I got 9/19 as well.

Approach:

How can a number be even?

Odd + odd and even + even

10/20 x 9/19 + 10/20 x 9/19

9/19
Director
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Re: Probability Q in Challenge 17 [#permalink]

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03 Jun 2005, 17:41
bmw wrote:
A Bin contains card numbered from 1 to 20. If two cards are drawn what is the probability the sum of the numbers on the cards is even ?

1/3
7/18
4/9
9/19
1/2

I think the answer should be 1/2 as opposed to 9/19 if am not wrong. For numbers 1-20, the sum of any 2 numbers is either even or odd, hence asked probability is 1/2

Can anyone point the flaw in the reasoning
thanks

prob= (2*(10c2))/20c2 = 2*9*5/19*10 = 9/19

as far as the flaw in the reasoning is concerned

even if odd+odd or even+even, we have 10 odd numbers and 10 even numbers
so total even possiblities is 10c2 + 10c2 = 45 + 45 = 90

odd if odd + even, total 10*10 = 100 possibilities

check 20c2 = 19*10 = 190 possibilities overall

different counting concepts
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03 Jun 2005, 23:33
We have 3 possible scenarios, the sum of O+E=O, E+E=E and O+O=E so the prob of the sum being even is 2/3
Senior Manager
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Re: Probability Q in Challenge 17 [#permalink]

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03 Jun 2005, 23:51
bmw wrote:
A Bin contains card numbered from 1 to 20. If two cards are drawn what is the probability the sum of the numbers on the cards is even ?

1/3
7/18
4/9
9/19
1/2

I think the answer should be 1/2 as opposed to 9/19 if am not wrong. For numbers 1-20, the sum of any 2 numbers is either even or odd, hence asked probability is 1/2

Can anyone point the flaw in the reasoning
thanks

I problem with your reasoning is that you pick both the numbers together and not one after the another, which is not bad, but you should conditional probability in that case , in the absence of which , it is akin to saying that the numbers can be replaced.

HMTG.
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Re: Probability Q in Challenge 17 [#permalink]

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05 Jun 2005, 11:53
bmw wrote:
A Bin contains card numbered from 1 to 20. If two cards are drawn what is the probability the sum of the numbers on the cards is even ?

1/3
7/18
4/9
9/19
1/2

I think the answer should be 1/2 as opposed to 9/19 if am not wrong. For numbers 1-20, the sum of any 2 numbers is either even or odd, hence asked probability is 1/2

Can anyone point the flaw in the reasoning
thanks

If the first number is odd, second number has to be odd and if the first number is even, second number has to be even.

if you've already picked an odd number out of the box the prob of getting an odd one on the next go is 9/19 (as you have one less odd card in the box).

same thing if you pick an even number...

thus since the prob of picking an odd or even at the beginning is 1/2 and 1/2

the answer is 1/2* 9/19 + 1/2*9/19 = 9/19
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Re: Probability Q in Challenge 17 [#permalink]

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05 Jun 2005, 13:03
2C(10,2)/C(20,2)=2*10*9/20*19=9/19

Very curiously 9/19<1/2
It is because after you pick your first card, say it is odd, then it is a little less likely that your second draw will give you another odd, since there are more even cards then odd cards. So it is always less likely that you will draw two odd cards, or two even card, and it is more likely that you will draw one of each type.
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Re: Probability Q in Challenge 17   [#permalink] 05 Jun 2005, 13:03
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