Probability question - Need expert help

What you did looked fine to me. I guess from the wording of the question, it's unclear to me whether the contestant wins by choosing 2 black marbles. That would make it very, very likely that the contestant would win, and that doesn't sound much like a Las Vegas game of chance to me. One other way to look at the problem:

the probability of picking, say, Blue AND Blue is (1/100)(1/100) = 1/10,000

and the same is true for each of the 5 unique colours. So the probability of picking a pair of the unique colours will be 5/10,000. Now, if we also need to consider the case of picking two black marbles, the probability that happens is (95/100)(95/100) = 95^2/10,000. Adding that to 5/10,000 gives the answer (the same answer you got above).

Now if your question was actually about how to simplify your answer, I suppose you could just long-multiply and it shouldn't take too long. To calculate larger squares, you can also take advantage of the difference of squares pattern, by finding a product of two numbers near to 95 that's easy to work with. From the difference of squares:

(95 - 5)(95 + 5) = 95^2 - 25

but we can see that the left side, (95 - 5)(95 + 5) is also equal to 90*100= 9000. So

9000 = 95^2 - 25
9025 = 95^2

and if we are supposed to count the selection of 2 black coloured balls as a 'win', then the answer is 9025/10,000 + 5/10,000 = 9030/10,000 = 903/1000.

voodoo_child wrote:Thanks Ian for helping me. I have two questions:-

#1- what's wrong with method #1 ?
#2 - OA for this problem is 5/10000 = 1/2000. The source of this problem is Veritas Prep. I believe that the author is assuming that only 5 different color pairs could be formed. Otherwise, I don't see how one can get that answer. (I don't have other answer choices.)

Thanks again for your help. It will be great if you could elaborate why method#1 doesn't work. I think that will be a good take away for me.

Thanks
Voodoo

I didn't notice your 'method 1' the first time. It's the numerator that isn't right; I don't know how you arrived at "total pairs = 100", but it's not the right number. The numerator in your 'method 2' had the right number of pairs in the numerator.

Clearly the OA is not considering a pair of black marbles to be a winning combination. I don't think the question makes clear whether that pair should be counted, so I don't care for the wording much.