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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, 06:09
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Question Stats:

65% (00:52) correct 35% (01:22) wrong based on 190 sessions

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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, 06:45, 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, 06:27
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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, 06:48
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, 09:26
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, 04:07
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, 12:55
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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, 11:39
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]

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28 Jun 2017, 13:22
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.

Hello Bunuel and VeritasPrepKarishma

As per Veritas the answer is 2/7

And the logical argument is that event is happening simultaneously hence we will only consider one scenario of removing the balls.

Could you please confirm what should be the correct strategy and what is the correct answer?

We are now confused between 4/7 & 2/7
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Re: A jar contains 4 black and 3 white balls. If you pick two ba [#permalink]

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28 Jun 2017, 13:33
ydmuley wrote:
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.

Hello Bunuel and VeritasPrepKarishma

As per Veritas the answer is 2/7

And the logical argument is that event is happening simultaneously hence we will only consider one scenario of removing the balls.

Could you please confirm what should be the correct strategy and what is the correct answer?

We are now confused between 4/7 & 2/7

Mathematically the probability of picking two balls simultaneously, or picking them one at a time (without replacement) is the same. The correct answer is 4/7.
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Re: A jar contains 4 black and 3 white balls. If you pick two ba [#permalink]

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28 Jun 2017, 13:40
Hello Bunuel and VeritasPrepKarishma

As per Veritas the answer is 2/7

And the logical argument is that event is happening simultaneously hence we will only consider one scenario of removing the balls.

Could you please confirm what should be the correct strategy and what is the correct answer?

We are now confused between 4/7 & 2/7[/quote]
Bunuel wrote:

Mathematically the probability of picking two balls simultaneously, or picking them one at a time (without replacement) is the same. The correct answer is 4/7.

Ok.. thanks Bunuel - will go with 4/7

VeritasPrepKarishma - Not sure if you agree to this, in case the books have to be changed to avoid confusion going forward.
_________________

"Nothing in this world can take the place of persistence. Talent will not: nothing is more common than unsuccessful men with talent. Genius will not; unrewarded genius is almost a proverb. Education will not: the world is full of educated derelicts. Persistence and determination alone are omnipotent."

Worried About IDIOMS? Here is a Daily Practice List: https://gmatclub.com/forum/idiom-s-ydmuley-s-daily-practice-list-250731.html#p1937393

Best AWA Template: https://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html#p470475

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

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29 Jun 2017, 04:56
ydmuley wrote:
Hello Bunuel and VeritasPrepKarishma

As per Veritas the answer is 2/7

And the logical argument is that event is happening simultaneously hence we will only consider one scenario of removing the balls.

Could you please confirm what should be the correct strategy and what is the correct answer?

We are now confused between 4/7 & 2/7

Bunuel wrote:

Mathematically the probability of picking two balls simultaneously, or picking them one at a time (without replacement) is the same. The correct answer is 4/7.

Ok.. thanks Bunuel - will go with 4/7

VeritasPrepKarishma - Not sure if you agree to this, in case the books have to be changed to avoid confusion going forward.

Responding to a pm:

The answer is certainly 4/7 and that is what the book says too.
The book shows that "2 simultaneous picks" is the same as "pick one and then another". So getting one black and one white can be achieved in 2 ways: a black and then a white or a white and then a black. I agree that the addition isn't explicitly shown but "pick two such that one is black and one is white" is composed of two cases:
First black and then white for which Probability = 2/7
First white and then black for which Probability = 2/7
They both result in one white and one black so answer would be 4/7.

Note that there are only 2 other cases:
Both black for which probability = 4/7 * 3/6 = 2/7
Both white for which probability = 3/7 * 2/6 = 1/7

Overall probability = 2/7 + 2/7 + 2/7 + 1/7 = 1
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A jar contains 4 black and 3 white balls. If you pick two ba   [#permalink] 29 Jun 2017, 04:56
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