There's a basket containing 10 apples - 7 red and 3 green. : PS Archive
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# There's a basket containing 10 apples - 7 red and 3 green.

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There's a basket containing 10 apples - 7 red and 3 green. [#permalink]

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30 May 2006, 09:58
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There's a basket containing 10 apples - 7 red and 3 green. Upon drawing 2 apples, What's the probability that we get 2 red and 1 green from next drawing of 3 apples?

it is a modified question from the question in the following link: http://www.gmatclub.com/phpbb/viewtopic.php?t=30016

No OA...
Senior Manager
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30 May 2006, 10:51
Professor wrote:
There's a basket containing 10 apples - 7 red and 3 green. Upon drawing 2 apples, What's the probability that we get 2 red and 1 green from next drawing of 3 apples?

it is a modified question from the question in the following link: http://www.gmatclub.com/phpbb/viewtopic.php?t=30016

No OA...

When you draw 2 apples, you get
1. 1 red and 1 green,
2. Both red,
3. Both green.

1. Probability of 1 red and 1 green
= C(7,1) * C(3,1) / C(10,2) = 21/45=7/15
After this, if we draw 3 more apples, the probability of getting 2R and 1G (from 6R and 2G)= C(6,2)*C(2,1)/C(8,3) = 30/56=15/28
Thus total of this branch = 7/15*15/27=1/4.
2. Probability of both red = C(7,2)*C(3,0)/C(10,2) = 21/45=7/15
Again, after this, if we draw 3 more apples, probability of getting 2R and 1G from 5R and 3G = C(5,2)*C(3,1)/C(8,3)=30/56 = 15/28
Thus total from this branch = 1/4.
3. Probability of both green = C(7,0)*C(3,2)/C(10,2) = 3/45=1/15.
Again, after this, if we draw 3 more apples, probability of getting 2R and 1G from 7R and 1G = C(7,2)*C(1,1)/C(8,3)=21/56.
Thus total probability from this branch = 7/280 = 1/40
Therefore total probability = 1/4+1/4+1/40=21/40.
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30 May 2006, 23:14
can you beleive it Prof, when I read the question on the link yesterday I was solving the question like the way this question has been framed.

Okay let me try

Two apples are drawn hence, both can be red or green or one can be red and one can be green,

--> WHen both are red.
Probability of getting both red = 7C2/10C2 --- eq 1
Probabilty of getting 2 red and 1 green afterwards = 5C2*3C1/8C3 ---eq 2
total P1 = eq1*eq2 ----------- eq3

--> when both green
Probability of getting both green = 2C2/10C2 --- eq 4
Probabilty of getting 2 red and 1 green afterwards = 7C2*1C1/8C3 ---eq 5
total P2 = eq4*eq5 --------------------eq6

--> when one is red and other is green
Probabibilty = 7C1*3C1/10C2 ------- eq 7
Probability of second drawing = 6C2*2C1/8C3 ---- eq8
Total P3 = eq7* eq8

Final P = P1 +p2 +p3
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31 May 2006, 04:52
There's a basket containing 10 apples - 7 red and 3 green. Upon drawing 2 apples, What's the probability that we get 2 red and 1 green from next drawing of 3 apples?

***********************************************

I came up with 21/40

From 7 red apples choose 2 (2C7) = 21 ---(1)

From 3 green apples choose 1 (1C3) = 1 ---(2)

We have 10 apples in total and we need to choose 3

3C10 = 120 ---(3)

Prob = ((1) + (2))/3

Prob = (21*3)/120 = 21/40
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31 May 2006, 16:01
Professor wrote:
There's a basket containing 10 apples - 7 red and 3 green. Upon drawing 2 apples, What's the probability that we get 2 red and 1 green from next drawing of 3 apples?

it is a modified question from the question in the following link: http://www.gmatclub.com/phpbb/viewtopic.php?t=30016

No OA...

good job Prof. Salute!!!
we need more of these prima facie scary problems. Looks like a very probable 700 + level question

7c2/10c2 (5c2x3c1/8c3) + 7c1c3c1(6c2x2c1/8c3) + 3c2(7c2x1c1/8c3)=21/40
Re: PS   [#permalink] 31 May 2006, 16:01
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