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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