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# A jar contains only black marbles and white marbles. If two thirds of

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Joined: 02 Sep 2009
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A jar contains only black marbles and white marbles. If two thirds of  [#permalink]

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21 Apr 2015, 05:59
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65% (hard)

Question Stats:

65% (01:53) correct 35% (06:49) wrong based on 216 sessions

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A jar contains only black marbles and white marbles. If two thirds of the marbles are black, how many white marbles are in the jar?

(1) If two marbles were to be drawn, simultaneously and at random, from the jar, there is a 5/12 probability that both would be black.

(2) If one white marble were removed from the jar, there would be a 1/4 probability that the next randomly-drawn marble would be white.

Kudos for a correct solution.

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A jar contains only black marbles and white marbles. If two thirds of  [#permalink]

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21 Apr 2015, 08:01
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Bunuel wrote:
A jar contains only black marbles and white marbles. If two thirds of the marbles are black, how many white marbles are in the jar?

(1) If two marbles were to be drawn, simultaneously and at random, from the jar, there is a 5/12 probability that both would be black.

(2) If one white marble were removed from the jar, there would be a 1/4 probability that the next randomly-drawn marble would be white.

Kudos for a correct solution.

1) probability of first blacck marble equal $$\frac{2}{3}$$. if we multiply this probability on probability second black marble, we receive common probability $$\frac{5}{12}$$
Let's x be second probability
$$\frac{2}{3} * x = \frac{5}{12}$$
$$x = \frac{5}{8}$$ and this probability after we remove one marble, so initially we have 9 marbles in total and $$\frac{1}{3}$$ of them white: 3 white marbles
Sufficient

2) let's x be total marbles
$$\frac{1}{3}x-1 = \frac{1}{4}(x-1)$$
$$x = 9$$ marbles in total and $$\frac{1}{3}$$ of them white: 3 white marbles
Sufficient.

Answer is D
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Re: A jar contains only black marbles and white marbles. If two thirds of  [#permalink]

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10 Mar 2016, 19:31
1
suppose we have X white, and 2X black
total 3X

now..
1 - probability that 2 black is:
2/3 * Y(variable) = 5/12
y=5/12*3/2
y=5/8
since y must be: (x-1)/(3x-1)
it must be true that there are 9 marbles, out of which 6 black and 2 white. sufficient

2. to draw a second white is 1/4
it must be true that x-1/3x-1 = 1/4
4x-1 = 3x-1
x=3.

sufficient.
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Joined: 14 Dec 2017
Posts: 477
Re: A jar contains only black marbles and white marbles. If two thirds of  [#permalink]

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03 Jul 2018, 10:06
Bunuel wrote:
A jar contains only black marbles and white marbles. If two thirds of the marbles are black, how many white marbles are in the jar?

(1) If two marbles were to be drawn, simultaneously and at random, from the jar, there is a 5/12 probability that both would be black.

(2) If one white marble were removed from the jar, there would be a 1/4 probability that the next randomly-drawn marble would be white.

Kudos for a correct solution.

Let $$X$$ be the total # of marbles. hence we have $$\frac{2X}{3}$$ black marbles & $$\frac{X}{3}$$ white marbles.

The question prompt now translates to finding X.

Statement 1: When two marbles are drawn, the probability of both to be black = $$\frac{5}{12}$$

Hence, $${(2X/3)}/X$$ * $$((2X/3) - 1)/(X-1)$$ = $$\frac{5}{12}$$

We can solve this for X. Hence Statement 1 alone is sufficient.

Statement 2: A white ball is drawn & then a white ball is drawn, the probability of drawing a second white ball is $$\frac{1}{4}$$

Hence we, $$((X/3) - 1)/(X - 1) = 1/4$$

We can solve this for X. Hence statement 2 alone is sufficient.

Answer D.

Thanks,
GyM
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A jar contains only black marbles and white marbles. If two thirds of  [#permalink]

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10 Jul 2018, 08:47
1) We can say that we have one part of white balls per two parts of black balls. We want to find how much is one part.

The probability of having 2 black balls is:
$$\frac{5}{12}$$=$$\frac{2p}{3p}$$.$$\frac{(2p-1)}{(3p-1)}$$

We can cancel out the p in the first fraction $$\frac{2p}{3p}$$ so we can find p with the resulting equation. (p=3)

Sufficient.

2) We know that initially the probability of having a white ball is 1:3. To find the probability of having the second white ball, we need to subtract one to each. Let's write down our hypothesis of the initial amount of white balls vs total that meet the 1:3 probability. Then let's write down how it would look like after we removed one white ball and see if the resulting probability is 1:4:

1:3 => this would make it impossible to have a second white ball
2:6 => 1:5 ; not what we want
3:9 => 2:8 which is equivalent to 1:4; what we want

We needed to have 3 out of 9 white balls initially to meet this condition.

Sufficient.
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A jar contains only black marbles and white marbles. If two thirds of &nbs [#permalink] 10 Jul 2018, 08:47
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