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12 Days of Christmas 2024 - 2025 Competition with $40,000 of Prizes

A jar contains only red, blue and green marbles. If 2 marbles are selected at random, one after the other, with replacement, what is the probability that the first marble is red and the second is green?

(1) The ratio of red marbles to the combined total of blue and green marbles is 1:2, and the ratio of blue marbles to the combined total of red and green marbles is 1:5.

(2) The jar contains 8 red marbles.

 


This question was provided by GMAT Club
for the 12 Days of Christmas Competition

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Probability that the first marble is red and the second is green is p(r)*p(g). [As mutually exclusive]

We need to know the p(r) & p(g).

Stmt (1) The ratio of red marbles to the combined total of blue and green marbles is 1:2, and the ratio of blue marbles to the combined total of red and green marbles is 1:5.

From this stmt we can get the ration of r:b:g is we solve both the given rations which is 2:3:3.

Hence Stmt 1 is sufficient to give both the required probabilities.


Stmt (2) The jar contains 8 red marbles.

This give now data about the other color marbles. Hence we cannot find the probability. Insufficient.

So IMO A
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We need to find [r/(r+b+g)] * [g/(r+b+g)]

Statement 1

r/(b+g) = 1/2, we can infer that r/(r+b+g) = 1/3
b/(r+g) = 1/5, we can infer that b/(r+b+g) = 1/6

Now, g/(r+b+g) = 1 - r/(r+b+g) - b/(r+b+g) = 1/2

We have values for r/(r+b+g) and g/(r+b+g). SUFFICIENT

Statement 2

r = 8, this doesn't give us any data on total no. of balls and green balls. NOT SUFFICIENT

Answer A.
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12 Days of Christmas 2024 - 2025 Competition with $40,000 of Prizes

A jar contains only red, blue and green marbles. If 2 marbles are selected at random, one after the other, with replacement, what is the probability that the first marble is red and the second is green?

(1) The ratio of red marbles to the combined total of blue and green marbles is 1:2, and the ratio of blue marbles to the combined total of red and green marbles is 1:5.

(2) The jar contains 8 red marbles.

 


This question was provided by GMAT Club
for the 12 Days of Christmas Competition

Win $40,000 in prizes: Courses, Tests & more

 


Probability = \(\frac{R }{ R + G + B} * \frac{G }{ R + G + B}\)

(1) The ratio of red marbles to the combined total of blue and green marbles is 1:2, and the ratio of blue marbles to the combined total of red and green marbles is 1:5.

\(\frac{R }{ B + G} = \frac{1}{2}\)

\(\frac{B }{ R + G} = \frac{1}{5}\)

\(2R = B + G\)

\(R = \frac{B + G }{ 2}\)

\(B = \frac{R + G }{5}\)

\(R = \frac{\frac{R + G}{5} + G}{2}\)

\(10R = R + 6G\)

\(9R = 6G\)

\(R = \frac{2}{3} G \)

Therefore we can express all the terms in a single variable and find a value for the required probability. Statement 1 is sufficient.

(2) The jar contains 8 red marbles.

Not sufficient to find the required probability.

Option A
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The probability in question is \( \frac{r}{T} * \frac{g}{T} * \frac{1}{2 }= \frac{rg}{2T^2} \) (I'm dividing in half to eliminate the cases where green is taken first). T stands for total (b+r+g).

(1)
  • \(\frac{r}{b+g}=\frac{1}{2}\)
\(2r=b+g\)
\(b=2r-g\)

  • \(\frac{b}{r+g}=\frac{1}{5} \)
\(5b=r+g\)
\(5(2r-g)=r+g\)
\(10r-5g=r+g\)
\(9r=6g, g=\frac{3r}{2}\)

Then total \(T = r+b+g=r+(2r-g)+g=3r\), and \(rg=r*\frac{3r}{2}=\frac{3r^2}{2}\)
And the probability:
\(\frac{rg}{2T^2}= \frac{3r^2}{2} / (3r)^2=\frac{3r^2}{2} / 9r^2\)
With cancelling out \(r^2\), we will arrive at the answer. Sufficient.

Hence, condition 2 is not needed (and of course insufficient alone), and the answer is A.
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