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Six cards numbered from 1 to 6 are placed in an empty bowl.

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Six cards numbered from 1 to 6 are placed in an empty bowl.

by mitzwillrockgmat » Mon May 24, 2010 11:36 am
Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a second card is drawn. If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5 ?

A. 1/6
B. 1/5
C. 1/3
D. 2/5
E. 2/3

the only ways to get sum 8 is if we pick

2 ,6 or 6 ,2
3, 5 or 5, 3
4, 4 or 4, 4

I want to know why we don't count 4, 4 twice? this is essentially a combination problem so why not total combinations 6 rather than 5?

favorable outcome will remain 2 either ways 3, 5 or 5,3.

Please explain :)
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by Patrick_GMATFix » Mon May 24, 2010 11:40 am
4,4 and 4,4 doesn't get counted as distinct outcomes because they are exactly the same (we're not picking two cards each numbered 4, but the same card). In each case, the same card is selected, put bag, and selected again.

This is different from {3,5} and {5,3}. In the latter scenario, the two events are different: a different card is picked first each time.

Hope that helps
-Patrick
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by jeffedwards » Mon May 24, 2010 12:32 pm
What is the OA? I got [spoiler]D- 2/5[/spoiler]

The way I went about doing this is as follows.

The person drawing could pick any number 2-6 inclusive (to add up to 8). Then I listed the possible combos assuming the numbers add up to 8. For each number there is only one other number that the person can choose.

2-6
3-5
4-4
5-3
6-2

Then I counted the number total sets and sets with a 5.... got [spoiler]2/5[/spoiler]
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by KKGMATTER » Mon May 24, 2010 8:44 pm
Please post the OA
Im getting 1/3 from 2/6
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by nxanand » Tue May 25, 2010 12:12 am
I got 2/5 (D)

no.of ways to get the sum of 8 = 5 : (2,6),(6,2),(3,5),(5,3),(4,4)

no.of ways that one card is 5 = 2 : (3,5),(5,3)

therefore, probability = 2/5
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by sanju09 » Tue May 25, 2010 12:15 am
Patrick_GMATFix wrote:4,4 and 4,4 doesn't get counted as distinct outcomes because they are exactly the same (we're not picking two cards each numbered 4, but the same card). In each case, the same card is selected, put bag, and selected again.

This is different from {3,5} and {5,3}. In the latter scenario, the two events are different: a different card is picked first each time.

Hope that helps
-Patrick
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