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=30016No 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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