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bagdbmba
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A jar contains 4 black and 3 white balls. If you pick two ba Updated on: 30 Aug 2013, 06:45
Originally posted by bagdbmba on 30 Aug 2013, 06:09.
Last edited by bagdbmba on 30 Aug 2013, 06:45, edited 1 time in total.
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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69% (01:09) correct 31% (01:33) wrong based on 497 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
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C
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Bunuel
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A jar contains 4 black and 3 white balls. If you pick two ba 30 Aug 2013, 06:27
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bagdbmba
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
Show more

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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A jar contains 4 black and 3 white balls. If you pick two ba 30 Aug 2013, 06:48
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Thanks Bunuel for pointing it out...Corrected the OA accordingly in the question...
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A jar contains 4 black and 3 white balls. If you pick two ba 05 Sep 2013, 09:26
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Bunuel
bagdbmba
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.
Show more

Hi Bunuel,
As per the above explanation - why the OA of the following problem is NOT 3/10?
https://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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A jar contains 4 black and 3 white balls. If you pick two ba 08 Sep 2013, 04:07
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Bunuel
bagdbmba
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.
Show more

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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A jar contains 4 black and 3 white balls. If you pick two ba 28 Jun 2017, 13:22
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Bunuel
bagdbmba
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.
Show more

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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A jar contains 4 black and 3 white balls. If you pick two ba 28 Jun 2017, 13:33
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ydmuley
Bunuel
bagdbmba
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.

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
Show more

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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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


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.
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A jar contains 4 black and 3 white balls. If you pick two ba 29 Jun 2017, 04:56
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ydmuley
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
Show more
Bunuel


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.
Show more

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 29 Oct 2021, 08:35
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Probability of both balls being Black = 4/7*3/6 = 2/7,
Probability of both balls being White = 3/7* 2/6= 1/7
Probability of 1 white one black = 1- (2/7+17) = 4/7
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A jar contains 4 black and 3 white balls. If you pick two ba 28 Jan 2026, 10:23
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Deconstructing the Question

Method: Combinations (Selection)

1. Total Outcomes (Denominator)
Total balls = \(4 (Black) + 3 (White) = 7\).
We select 2 balls.
\(Total = 7C2 = (7*6)/2 = \) 21

2. Favorable Outcomes (Numerator)
We need exactly 1 Black AND 1 White.
* Ways to pick 1 Black from 4: \(4C1 = 4\)
* Ways to pick 1 White from 3: \(3C1 = 3\)

\(Favorable = 4 * 3 = \) 12

3. Probability
\(P = Favorable / Total\)
\(P = 12 / 21\)
Simplify by dividing by 3:
\(P = \) 4 / 7

The correct answer is C.
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