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# A barrel contains only red balls, white balls, and brown balls. If two

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A barrel contains only red balls, white balls, and brown balls. If two  [#permalink]

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Updated on: 24 Oct 2017, 21:59
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A barrel contains only red balls, white balls, and brown balls. If two balls are selected at random without replacing each ball, what is the probability that two brown balls will be selected?

(1) The probability that one ball selected at random is either red or white is $$\frac{2}{3}$$

(2) There are nine balls in the barrel.

Originally posted by gauravsoni on 20 May 2014, 12:39.
Last edited by Bunuel on 24 Oct 2017, 21:59, edited 3 times in total.
Edited the question.
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Re: A barrel contains only red balls, white balls, and brown balls. If two  [#permalink]

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20 May 2014, 20:47
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gauravsoni wrote:
A barrel contains only red balls, white balls, and brown balls. If two balls are selected at random without replacing each ball, what is the probability that two brown balls will be selected?

(1) The probability that one ball selected at random is either red or white is .

(2) There are nine balls in the barrel.

This is an interesting one. The question says we have only red, white and brown balls. And we have to find the probability that both balls drawn with replacement are brown.

Probability of getting 1 brown ball = number of brown balls/ total number of balls = A

Probability that both balls drawn are brown = A.A = A^2. (Note that A is dependent on number of brown balls and total number of balls which are 2 variables)

Statement 1: The probability that one ball selected at random is either red or white is 2/3.

This means probability of not getting a brown ball =2/3
Therefore, A which is probability of getting a brown ball is 1/3

Now since, we want to get A^2, it may seem that we can answer the question. But this is a trap. Lets say, we have only 3 balls. A=1/3 means, we have only 1 brown ball. Thus, probability of getting 2 brown balls become 0. In case it is more than 3 balls, i.e. 6, 9, 12, ans so on, we can predict the probability.
Hence, this is insufficient

Statement 2: There are nine balls in the barrel.

We don't know the number of brown balls and hence can't predict the probability.

Hence insufficient.

Now, let us take both the statement together. After statement 1, if we can assure that the barrel has more than 3 balls, we can definitely tell the probability. Now, we are given 9 balls. Which means we can calculate the probability.

Hence, answer is C.

Hope it helps!!

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Re: A barrel contains only red balls, white balls, and brown balls. If two  [#permalink]

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05 Sep 2018, 10:54
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gauravsoni wrote:
A barrel contains only red balls, white balls, and brown balls. If two balls are selected at random without replacing each ball, what is the probability that two brown balls will be selected?

(1) The probability that one ball selected at random is either red or white is $$\frac{2}{3}$$

(2) There are nine balls in the barrel.

Statement One Alone:

The probability that one ball selected at random is either red of white is 2/3.

That means the probability that one ball selected at random is brown is 1/3. However, since we don’t know how many brown balls there are, we can’t determine the probability that two brown balls will be selected without replacement. For example, if there are 2 brown balls in the barrel (before any selection), the probability that two brown balls will be selected without replacement is 2/6 x 1/5 = 1/3 x 1/5 = 1/15. However, if there are 3 brown balls in the barrel (before any selection), the probability that two brown balls will be selected without replacement is 3/9 x 2/8 = 1/3 x 1/4 = 1/12.

Statement one alone is not sufficient.

Statement Two Alone:

There are nine balls in the barrel.

Without knowing the number of balls of each color (especially the color brown) or the probability of selecting a certain color (especially the color brown), we can’t determine the probability that two brown balls will be selected without replacement.

Statement two alone is not sufficient.

Statements One and Two Together:

With the two statements, we can determine that there are 3 brown balls in the barrel (before any selection) since 1/3 x 9 = 3. Knowing that there are 3 brown balls in the barrel (before any selection), we can also determine the probability that two brown balls will be selected without replacement. That probability is is 3/9 x 2/8 = 1/3 x 1/4 = 1/12.

Answer: C
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Re: A barrel contains only red balls, white balls, and brown balls. If two  [#permalink]

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20 May 2014, 21:41
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gauravsoni wrote:
A barrel contains only red balls, white balls, and brown balls. If two balls are selected at random without replacing each ball, what is the probability that two brown balls will be selected?

(1) The probability that one ball selected at random is either red or white is $$\frac{2}{3}$$

(2) There are nine balls in the barrel.

Can someone please explain the working ?

Let there be x red balls, y, white balls and z brown balls .

From st 1 we have that the probability of selection either red or white is 2/3

or (x+y)/ (x+y+z)= 2/3 -----We see that x+y=2z or total no. of balls =3z

So probability of drawing 2 brown balls will be $$zc2/3zc2$$ or z(z-1)/(3z-1)*(3z-2)------Thus till we know z we can find the probability

St 2 says that x+y+z=9 so not sufficient

Combining we see that 3z=9 and z=3 so probability will be 3/9 or 1/3.

Ans is C.....
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Re: A barrel contains only red balls, white balls, and brown balls. If two  [#permalink]

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21 May 2014, 02:27
1
A barrel contains only red balls, white balls, and brown balls. If two balls are selected at random without replacing each ball, what is the probability that two brown balls will be selected?

(1) The probability that one ball selected at random is either red or white is $$\frac{2}{3}$$. This implies that 1/3rd of the balls in the barrel are brown. If there are only 3 balls in the barrel then the probability of selecting two brown balls is obviously 0 but if there are say 6 balls in the barrel then the probability of selecting two brown balls is greater than 0. Not sufficient.

(2) There are nine balls in the barrel. Clearly insufficient.

(1)+(2) 1/3*9=3 balls out of 9 are brown. We can get the probability. Sufficient.

Answer: C.

Remember, on DS problems, all you need to do is evaluate whether you would be able to arrive at the answer using the information provided in each statement; you don’t need to waste time actually finding the answer.
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Re: A barrel contains only red balls, white balls, and brown balls. If two  [#permalink]

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21 May 2014, 02:36
WoundedTiger wrote:
gauravsoni wrote:
A barrel contains only red balls, white balls, and brown balls. If two balls are selected at random without replacing each ball, what is the probability that two brown balls will be selected?

(1) The probability that one ball selected at random is either red or white is $$\frac{2}{3}$$

(2) There are nine balls in the barrel.

Can someone please explain the working ?

Let there be x red balls, y, white balls and z brown balls .

From st 1 we have that the probability of selection either red or white is 2/3

or (x+y)/ (x+y+z)= 2/3 -----We see that x+y=2z or total no. of balls =3z

So probability of drawing 2 brown balls will be $$zc2/3zc2$$ or z(z-1)/(3z-1)*(3z-2)------Thus till we know z we can find the probability

St 2 says that x+y+z=9 so not sufficient

Combining we see that 3z=9 and z=3 so probability will be 3/9 or 1/3.

Ans is C.....

We have 3 brown balls out of 9. The probability of selecting 2 brown balls is 3/9*2/8 = 1/12.
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A barrel contains only red balls, white balls, and brown balls. If two  [#permalink]

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29 Aug 2018, 12:23
gauravsoni wrote:
A barrel contains only red balls, white balls, and brown balls. If two balls are selected at random without replacing each ball, what is the probability that two brown balls will be selected?

(1) The probability that one ball selected at random is either red or white is $$\frac{2}{3}$$

(2) There are nine balls in the barrel.

$$?\,\,\,\, = \,\,\,P\left( {2\,\,br\,\,{\text{out}}\,\,{\text{of}}\,\,2\,\,{\text{sequential}}\,\,{\text{extractions}}\,\,{\text{no}}\,\,{\text{replacements}}} \right)\,\,$$

$$\left( 1 \right)\,\,P\left( {re\,\,{\text{or}}\,\,wh\,\,{\text{out}}\,\,{\text{of}}\,\,1\,\,{\text{extraction}}} \right) = \frac{2}{3}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,P\left( {br\,\,{\text{out}}\,\,{\text{of}}\,\,1\,\,{\text{extraction}}} \right) = \frac{1}{3}$$

$$\left\{ \begin{gathered} \,{\text{Take}}\,\,\,\left( {r,w,b} \right) = \left( {1,1,1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 0 \hfill \\ \,{\text{Take}}\,\,\,\left( {r,w,b} \right) = \left( {2,2,2} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,? \ne 0 \hfill \\ \end{gathered} \right.$$

$$\left( 2 \right)\,\,\,r + w + b\,\, = \,\,9$$

$$\left\{ \begin{gathered} \,{\text{Take}}\,\,\,\left( {r,w,b} \right) = \left( {7,1,1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 0 \hfill \\ \,{\text{Take}}\,\,\,\left( {r,w,b} \right) = \left( {6,1,2} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,? \ne 0 \hfill \\ \end{gathered} \right.$$

$$\left( {1 + 2} \right)\,\,\,br = \frac{1}{3}\left( 9 \right) = 3\,\,\,\,\, \Rightarrow \,\,\,\,\,?\,\, = \,\,{\text{unique}}\,\,\,\,\left( {\frac{3}{9} \cdot \frac{2}{8}} \right)$$

The above follows the notations and rationale taught in the GMATH method.
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A barrel contains only red balls, white balls, and brown balls. If two  [#permalink]

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02 Oct 2018, 21:07
HAd this been replacement allowed scenario. A would have been sufficient ?
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Re: A barrel contains only red balls, white balls, and brown balls. If two  [#permalink]

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24 Nov 2019, 03:26
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Re: A barrel contains only red balls, white balls, and brown balls. If two   [#permalink] 24 Nov 2019, 03:26
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