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## Probability and Combination Problems

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by vsr2014 » Tue Oct 21, 2014 9:59 am
beatthegmat wrote:Found this great set of probability and combination problems. Remember--these types of questions appear infrequently on the GMAT, so don't over-emphasize them too much in your studies!
#48 The solution provided doesn't seem to answer the right question. The question in the problem is "In order to ensure that the sum of all cards he drew was even, how many cards did Jerome have to draw?"
The solution says: The 12th draw ensures an even sum.
Let's assume Jerome drew 12 cards and there are 9 odd cards and 2 even cards, so the sum is ODD
If he drew 13 cards, 9 odds and 4 evens, the sum is still ODD.
SO he has to draw all 20 cards to ensure that the sum is EVEN.
I assume that he is not calculating the sum upon drawing every next card, because the condition doesn't say so..
Is my logic wrong here?

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by Rich.C@EMPOWERgmat.com » Tue Oct 21, 2014 10:53 am
Hi tbasebal24 and vsr2014,

The original post is over 2 years old and the response appears to be to a different question, but for anyone else who might be interested... here's how you can solve this problem.

In probability questions, you can either calculate what you "want" or what you "don't want"; certain probability questions are easier to deal with if you focus on what you "don't want" (since the two possibilities will always add up to 1).

Here, we have 3 red marbles and 7 blue marbles. We're told to select 2 marbles at random and we're asked for the probability that AT LEAST ONE will be blue. Instead of calculating all the different ways to get 1 or 2 blue marbles, let's calculate what we DON'T want: 0 blue marbles.

On the first pull, we'd have a 3/10 chance of NOT getting a blue marble. After pulling out that first marble....
On the second pull, we'd have a 2/9 chance of NOT getting a blue marble.

(3/10)(2/9) = 6/90 = 1/15 --> this is the probability of NOT getting what we want.

1 - 1/15 = 14/15 --> this is the probability of getting what we WANT.