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Probability question - Need expert help

by voodoo_child » Tue Aug 28, 2012 5:08 pm
In a $10 game of chance in Las Vegas, there are two identical bowls that contain 100 marbles. Each bowl contains black colored 95 marbles along with 5 different colored marbles: one blue, one yellow, one green, one red and one gold. You get to pick one marble randomly from each bowl. If you get a matching pair of colored marbles (blue-blue, gold-gold), you win $10000. What is the probability that you win the prize in one play of the game?

HEre's what I tried :

Method 1 = total pairs = 100; Total combinations = 100*100 => PRob = 1/100

Method 2 (really crash landed) --

Choose 1 of the black colored balls from the first bowl and then the second one => 95C1 * 95C1
Choose 1 of blue balls => 1C1 * 1C1 ...Now this will be done for all the five colors. Hence, total = 5*1c1 * 1c1 = 5

Total = (95^2 + 5)/(100^2) = CRASHED !

Can any of the experts please help me ?


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by Ian Stewart » Tue Aug 28, 2012 6:20 pm
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.
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by voodoo_child » Tue Aug 28, 2012 7:23 pm
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
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by Ian Stewart » Tue Aug 28, 2012 7:47 pm
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.
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by das.ashmita » Tue Aug 28, 2012 9:44 pm
This is how I tried it:

Probability of choosing a colored marble from bowl 1 = 5/100 = 1/20
Probability of choosing the same colored marble from bowl 2 = 1/100

Probability (Win) = (1/20)(1/100) = 1/2000
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by voodoo_child » Wed Aug 29, 2012 5:55 am
Ian Stewart wrote: 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.
Thanks Ian. When I got up this morning, I realized why Method1 is incorrect. There cannot be hundred color pairs because there are two different bowls that each contains 95 black balls. There itself, the number of pairs = 95*95. Just '95' is incorrect.

On a separate note, I have a SC question. Which one is correct?
A) two different bowls that each contains
Or
B) two different bowls that each contain?

I am a bit confused. My intuition says A) because "each" is singular.
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by GaneshMalkar » Fri Aug 31, 2012 6:45 am
Assuming that black-black combination wont win any thing....

From Bowl1 we have 5 different colors of balls...so 5 ways of selecting one ball
as from second bowl we need to take the ball of same color so only 1 way.
Total way of win for one combination = 5 * 1
P(win) = 5/10000 = 1/2000
If you cant explain it simply you dont understand it well enough!!!
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by mohan514 » Fri Aug 31, 2012 8:58 am
the chances of getting one of the coloured ball in the fiest ball is 5/100

to win he should pick up the ball of same colour from the second bowl and that should be of the same colour

so sinjce there are only 5 coloured balls and among them only one would match the requirement

hence the chances are 1/100

so both should happen at once
hence on multiplication we reach 0.0005

is it right??
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