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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] New post 30 Aug 2013, 05:09
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A
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C
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Question Stats:

59% (01:42) correct 40% (01:16) wrong based on 52 sessions
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] New post 30 Aug 2013, 05:27
Expert's post
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.

Answer: C.

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] New post 30 Aug 2013, 05:48
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Re: A jar contains 4 black and 3 white balls.If you pick two [#permalink] New post 05 Sep 2013, 08:26
Expert's post
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.

Answer: C.

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

Would you please help?
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Re: A jar contains 4 black and 3 white balls.If you pick two [#permalink] New post 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.

Answer: C.

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   [#permalink] 08 Sep 2013, 03:07
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