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A box contains red balls, white balls, and yellow balls. If a ball is
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Updated on: 06 Feb 2015, 12:12
Question Stats:
88% (00:32) correct 12% (00:48) wrong based on 120 sessions
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A box contains red balls, white balls, and yellow balls. If a ball is randomly selected, what is the probability that it will NOT be yellow? (1) The probability that the ball will be red is 1/2. (2) The probability that the ball will be yellow is 1/3. *Nice one  I think. It is not difficult, but has a tiny trick if you are not careful.
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Originally posted by pacifist85 on 06 Feb 2015, 11:24.
Last edited by Bunuel on 06 Feb 2015, 12:12, edited 1 time in total.
Edited the question.



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Re: A box contains red balls, white balls, and yellow balls. If a ball is
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06 Feb 2015, 15:52
pacifist85 wrote: A box contains red balls, white balls, and yellow balls. If a ball is randomly selected, what is the probability that it will NOT be yellow?
(1) The probability that the ball will be red is 1/2. (2) The probability that the ball will be yellow is 1/3. *Nice one  I think. It is not difficult, but has a tiny trick if you are not careful. Dear pacifist85, This appears to be a very straightforward test of the complement ruleP(not A) = 1  P(A)Consider " the probability of that A happens" and " the probability that A doesn't happen"  having either one automatically gives us the other. See: http://magoosh.com/gmat/2012/gmatmath ... question/Is the complement rule the "trick" you had in mind? Mike
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Re: A box contains red balls, white balls, and yellow balls. If a ball is
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06 Feb 2015, 16:03
mikemcgarry wrote: pacifist85 wrote: A box contains red balls, white balls, and yellow balls. If a ball is randomly selected, what is the probability that it will NOT be yellow?
(1) The probability that the ball will be red is 1/2. (2) The probability that the ball will be yellow is 1/3. *Nice one  I think. It is not difficult, but has a tiny trick if you are not careful. Dear pacifist85, This appears to be a very straightforward test of the complement ruleP(not A) = 1  P(A)Consider " the probability of that A happens" and " the probability that A doesn't happen"  having either one automatically gives us the other. See: http://magoosh.com/gmat/2012/gmatmath ... question/Is the complement rule the "trick" you had in mind? Mike Hi Mike, Indeed the "1minus the opposite" was mainly the trick I was referring to. But also, I think that you can get confused because the problem does not actually give you the number of the balls  so we don't know how many they are. So, initially, one might think that, oh, none of the 2 data gives me the number of balls, so I cannot possibly answer this question. Then, being in a hurry anyway, you may disregard the second data point and move on. Also, the first data point gives you 1/2, which is 50% and the second one refers to the second of the 3 colours. This may incorrectly lead you to assume that you know how many half of the balls are. The rest is the other 50%. However, we have three colours here and the data points only refer to 2. Your poor, tired eye, having seen the 50% and reading clues about 2 of the three colours, might forget that the question stem actually mentioned 3 colours. Or is it only me...



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Re: A box contains red balls, white balls, and yellow balls. If a ball is
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06 Feb 2015, 16:15
pacifist85 wrote: Hi Mike, Indeed the "1minus the opposite" was mainly the trick I was referring to. But also, I think that you can get confused because the problem does not actually give you the number of the balls  so we don't know how many they are. So, initially, one might think that, oh, none of the 2 data gives me the number of balls, so I cannot possibly answer this question. Then, being in a hurry anyway, you may disregard the second data point and move on. Also, the first data point gives you 1/2, which is 50% and the second one refers to the second of the 3 colours. This may incorrectly lead you to assume that you know how many half of the balls are. The rest is the other 50%. However, we have three colours here and the data points only refer to 2. Your poor, tired eye, having seen the 50% and reading clues about 2 of the three colours, might forget that the question stem actually mentioned 3 colours. Or is it only me... Dear pacifist85, The magic of the complement rule and the other probability rules is they absolutely do not depend on knowing the number of items involved. In its essence, probability is a ratio, so ratio information (which includes percents) suffices for calculating probabilities. If we happen to know exact counts, we also can find ratios from those, but understand that sort of information is secondary. I think the way you are describing it is idiosyncratic to you. They are asking about yellow, and statement 2 tells me about yellow. Done. Does this make sense? Mike
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Re: A box contains red balls, white balls, and yellow balls. If a ball is
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06 Feb 2015, 16:35
mikemcgarry wrote: pacifist85 wrote: Hi Mike, Indeed the "1minus the opposite" was mainly the trick I was referring to. But also, I think that you can get confused because the problem does not actually give you the number of the balls  so we don't know how many they are. So, initially, one might think that, oh, none of the 2 data gives me the number of balls, so I cannot possibly answer this question. Then, being in a hurry anyway, you may disregard the second data point and move on. Also, the first data point gives you 1/2, which is 50% and the second one refers to the second of the 3 colours. This may incorrectly lead you to assume that you know how many half of the balls are. The rest is the other 50%. However, we have three colours here and the data points only refer to 2. Your poor, tired eye, having seen the 50% and reading clues about 2 of the three colours, might forget that the question stem actually mentioned 3 colours. Or is it only me... Dear pacifist85, The magic of the complement rule and the other probability rules is they absolutely do not depend on knowing the number of items involved. In its essence, probability is a ratio, so ratio information (which includes percents) suffices for calculating probabilities. If we happen to know exact counts, we also can find ratios from those, but understand that sort of information is secondary. I think the way you are describing it is idiosyncratic to you. They are asking about yellow, and statement 2 tells me about yellow. Done. Does this make sense? Mike Thank you. Yes it does make sense. DS questions are my weak point, exactly because I mix the 2 data, or just start actually solving them. I guess this is why I posted it. Not because the problem is difficult itself, but because the question is confusing me.



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