A container contains 4 red marbles and 8 blue marbles. A second contai

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

Answer: C

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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