Bunuel wrote:
A jar has 10 marbles, a mix of red and white. Two marbles are randomly chosen from the jar. If b is the probability that both will be red, is b > 1/3?
(1) Less than 1/2 of the marbles in the jar are white.
(2) The probability that 1 white marble and 1 red marble will be chosen together is 7/15.
Kudos for a correct solution.
I like spending a bit more time on the question stem with something like this, so that I don't have to do a bunch of calculations twice (once for each statement).
Let's say that the number of red marbles is R.
Then, what's the probability that both marbles are red?
The probability that the first is red is R/10, and the probability that the second is red is (R-1)/9.
So, the overall probability is R(R-1) / 90.
Therefore, the question is really asking: Is R(R-1) / 90 > 1/3?
Simplify: Is R(R-1) > 30?
When will this happen? Well, I know that 6*5 is exactly 30. So, if R is bigger than 6, the answer will be "yes". But if R is 6 or smaller, the answer will be "no".
The question really says,
Is R greater than 6?Statement 1:
Insufficient - R is greater than 5. So, there could be 6 red marbles ('no') or 9 red marbles ('yes').
Statement 2: I'm going to say this is
insufficient without doing any calculation. This is a safe assumption because of how probability works. Let's say that we had 3 red marbles and 7 white marbles. We'd have the same probability of getting one red and one white, as we would if we had 3 white marbles and 7 red marbles. So, just from this probability, you can't tell if there are a lot of red marbles, or only a few. Switching the number of red with the number of white will give you the same probability.
Statements 1 and 2: This seems sufficient for the same reason described in statement 2 above. Since we now know that R is greater than 5, we can no longer 'switch' the numbers of red and white marbles to end up with the same probability.
But if you want to do the calculation, here's how!
7/15 = probability of one red, one white
7/15 = probability of red first, white second + probability of white first, red second
7/15 = R/10 * (10-R)/9 + (10-R)/10 * R/9 = 2*R*(10-R)/90
Multiply both sides by 90:
42 = 2*R*(10-R)
21 = R*(10-R)
R could be 7 or 3. But, 3 is excluded by statement 1, so R can only be 7, and the answer is "yes". The statements together are
sufficient.
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