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 ?
Thanks
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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.
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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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
#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
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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.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
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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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
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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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.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.
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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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
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!!!
- Genius
- Genius
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??
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??