V1

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V1

by oquiella » Wed Nov 04, 2015 6:23 am
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.


Please breakdown the math

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by MartyMurray » Wed Nov 04, 2015 11:31 pm
We know that 2/3 of the marbles are black. This could be the case with various numbers of marbles in the jar.

For instance, there could be 2 black marbles and 1 white marble. Alternatively there could be 60 black marbles and 30 white marbles.

Now let's look at the statements.

Statement 1 tells us the probability of drawing two black marbles. Even though the statement says that they are to be drawn simultaneously, when doing this kind of calculation we act as if one were drawn before the other.

Let's try a couple. If there were 2 black and 1 white in the jar, the probability of drawing two black marbles would be 2/3 x 1/2 = 1/3. So given the constraint in Statement 1, we know that there are not 2 black marbles and 1 white marble in the jar.

Now let's try 8 and 4. The probability of drawing 2 black ones would be 2/3 x 7/11 = 14/33. So 8 and 4 does not work either.

The truth is that I am illustrating a point here, and actually we don't have to do any math to determine whether Statement 1 is sufficient. The point is that each different combination of black and while marbles yields a different probability of drawing two black ones. So that 5/12 probability given by Statement 1 only works with a particular combination of numbers of black and white marbles. We could calculate the numbers but we don't have to. We know that there is a unique pair of numbers that fits and so Statement 1 is sufficient.

Statement 2 basically says that when in the jar there is one fewer white marble than needed to make a 2:1 ratio of black to while marbles, there is a 1/4 chance of drawing a white marble. Once again, there will be only one 2:1 set of numbers of black and white marbles that will yield this result. In this case that set is 6 black marbles and 3 white marbles, though knowing that does NOT matter. I am including it to illustrate how this works, and you can see that any other 2:1 combination of black to white marbles will not fit the constraint of Statement 2.

So Statement 2 is also sufficient.

The correct answer is D.
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