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# A jar contains 4 black and 3 white balls. If you pick two ba

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A jar contains 4 black and 3 white balls. If you pick two ba [#permalink]

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30 Aug 2013, 05:09
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A jar contains 4 black and 3 white balls. If you pick two balls at the same time, what's the probability that one ball is black and one is white?

A. 2/7
B. 5/7
C. 4/7
D. 3/7
E. 1/2
[Reveal] Spoiler: OA

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Last edited by bagdbmba on 30 Aug 2013, 05:45, edited 1 time in total.
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Re: A jar contains 4 black and 3 white balls.If you pick two [#permalink]

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30 Aug 2013, 05:27
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bagdbmba wrote:
A jar contains 4 black and 3 white balls. If you pick two balls at the same time, what's the probability that one ball is black and one is white?

A. 2/7
B. 5/7
C. 4/7
D. 3/7
E. 1/2

P(1st black, 2nd white) = 4/7*3/6 = 4/14;
P(1st white, 2nd black) = 3/7*4/6 = 4/14.

P = 4/14 + 4/14 = 4/7.

OR: $$P=\frac{C^1_4*C^1_3}{C^2_7}=\frac{4}{7}$$

OA is NOT correct.
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Re: A jar contains 4 black and 3 white balls. If you pick two ba [#permalink]

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30 Aug 2013, 05:48
Thanks Bunuel for pointing it out...Corrected the OA accordingly in the question...
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Re: A jar contains 4 black and 3 white balls.If you pick two [#permalink]

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05 Sep 2013, 08:26
Bunuel wrote:
bagdbmba wrote:
A jar contains 4 black and 3 white balls. If you pick two balls at the same time, what's the probability that one ball is black and one is white?

A. 2/7
B. 5/7
C. 4/7
D. 3/7
E. 1/2

P(1st black, 2nd white) = 4/7*3/6 = 4/14;
P(1st white, 2nd black) = 3/7*4/6 = 4/14.

P = 4/14 + 4/14 = 4/7.

OR: $$P=\frac{C^1_4*C^1_3}{C^2_7}=\frac{4}{7}$$

OA is NOT correct.

Hi Bunuel,
As per the above explanation - why the OA of the following problem is NOT 3/10?
http://gmatclub.com/forum/x-y-and-z-are-all-unique-numbers-if-x-is-chosen-randomly-159208.html

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Re: A jar contains 4 black and 3 white balls.If you pick two [#permalink]

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08 Sep 2013, 03:07
Bunuel wrote:
bagdbmba wrote:
A jar contains 4 black and 3 white balls. If you pick two balls at the same time, what's the probability that one ball is black and one is white?

A. 2/7
B. 5/7
C. 4/7
D. 3/7
E. 1/2

P(1st black, 2nd white) = 4/7*3/6 = 4/14;
P(1st white, 2nd black) = 3/7*4/6 = 4/14.

P = 4/14 + 4/14 = 4/7.

OR: $$P=\frac{C^1_4*C^1_3}{C^2_7}=\frac{4}{7}$$

OA is NOT correct.

Hi Bunuel

I wanted to know if probability approach can be used to resolve any sort of probability problem. I am quite comfortable with it. when is it advantageous to use combination approach?
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Re: A jar contains 4 black and 3 white balls. If you pick two ba [#permalink]

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16 Mar 2015, 11:55
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Re: A jar contains 4 black and 3 white balls. If you pick two ba [#permalink]

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06 Jul 2016, 10:39
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: A jar contains 4 black and 3 white balls. If you pick two ba   [#permalink] 06 Jul 2016, 10:39
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# A jar contains 4 black and 3 white balls. If you pick two ba

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