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Math Expert V
Joined: 02 Sep 2009
Posts: 59712
A container contains 4 red marbles and 8 blue marbles. A second contai  [#permalink]

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Difficulty:   65% (hard)

Question Stats: 56% (02:32) correct 44% (02:18) wrong based on 98 sessions

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A container contains 4 red marbles and 8 blue marbles. A second container contains 6 red marbles and x blue marbles. One marble is drawn from each of the two containers. If the probability of drawing a pair of marbles of the same color is 1/2, what is the value of x?

A. 0
B. 4
C. 6
D. 10
E. 12

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Intern  B
Joined: 19 May 2018
Posts: 14
Re: A container contains 4 red marbles and 8 blue marbles. A second contai  [#permalink]

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1
For pp to be 0.5 no of marbles have to be equal, therefore x =6

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Senior Manager  P
Joined: 18 Jan 2018
Posts: 304
Location: India
Concentration: General Management, Healthcare
Schools: Booth '22, ISB '21, IIMB
GPA: 3.87
WE: Design (Manufacturing)
Re: A container contains 4 red marbles and 8 blue marbles. A second contai  [#permalink]

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2 possibilities : RR and BB

P(RR) = 4/12 * 6/(6+x)
P(BB) = 8/12 * x(6+x)

total Probability =[ 4/12 * 6/(6+x) ] + [ 8/12 * x/(6+x) ] = 1/2 ( given)

Tried , test the answers : option c fits
Target Test Prep Representative V
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Re: A container contains 4 red marbles and 8 blue marbles. A second contai  [#permalink]

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Bunuel wrote:
A container contains 4 red marbles and 8 blue marbles. A second container contains 6 red marbles and x blue marbles. One marble is drawn from each of the two containers. If the probability of drawing a pair of marbles of the same color is 1/2, what is the value of x?

A. 0
B. 4
C. 6
D. 10
E. 12

We can create the equation:

P(1st container = red and 2nd container = red) + (1st container = blue and 2nd container = blue) = 1/2

4/12 * 6/(6 + x) + 8/12 * x/(6 + x) = 1/2

1/3 * 6/(6 + x) + 2/3 * x/(6 + x) = ½

2/(6 + x) + 2x/[3(6 + x)] = 1/2

Multiplying the equation by 6(6 + x), we have:

12 + 4x = 3(6 + x)

12 + 4x = 18 + 3x

x = 6

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Senior Manager  G
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Re: A container contains 4 red marbles and 8 blue marbles. A second contai  [#permalink]

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Bunuel wrote:
A container contains 4 red marbles and 8 blue marbles. A second container contains 6 red marbles and x blue marbles. One marble is drawn from each of the two containers. If the probability of drawing a pair of marbles of the same color is 1/2, what is the value of x?

A. 0
B. 4
C. 6
D. 10
E. 12

We can PLUG IN THE ANSWERS, which represent the value of x.
The probability of drawling 2 of the same color = P(both marbles are red) or P(both marbles are blue) = P(RR) + P(BB).
When the correct answer is plugged in, we get:
$$P(RR) + P(BB) = \frac{1}{2}$$

B: 4 --> 1st container = 4 red and 8 blue, 2nd container = 6 red and 4 blue
$$P(RR) + P(BB) = (\frac{4}{12} * \frac{6}{10}) + (\frac{8}{12} * \frac{4}{10}) = \frac{3}{15} + \frac{4}{15} = \frac{7}{15}$$
Here, the resulting probability is less than $$\frac{1}{2}$$and thus is TOO SMALL.

D: 10 --> 1st container = 4 red and 8 blue, 2nd container = 6 red and 10 blue
$$P(RR) + P(BB) = (\frac{4}{12} * \frac{6}{16} + (\frac{8}{12} * \frac{4}{16}) = \frac{3}{24} + \frac{10}{24} = \frac{13}{24}$$
Here, the resulting probability is greater than $$\frac{1}{2}$$ and thus is TOO BIG.

Since B yields a result that is TOO SMALL, while D yields a result that is TOO BIG, the correct answer must between B and D.

C: 6 --> 1st container = 4 red and 8 blue, 2nd container = 6 red and 6 blue
$$P(RR) + P(BB) = (\frac{4}{12} * \frac{6}{12}) + (\frac{8}{12} * \frac{6}{12}) = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2}$$
Success!
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Available for tutoring in NYC and long-distance. Re: A container contains 4 red marbles and 8 blue marbles. A second contai   [#permalink] 27 Jul 2019, 03:09
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