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Re M2527
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16 Sep 2014, 00:23
Official Solution: (1) 20% of all chips in the basket are green. Therefore 80% of all chips in the basket are red, which means that the ratio of the number of red chips to the number of green chips is 4:1 (80:20). Now, if there are total of 5 chips in the basket (4 red + 1 green) then the probability that both chips will be green will be 0 (since there are NOT two chips in a basket) but if there are total of 10 chips in the basket (8 red + 2 green) then the probability that both chips will be green will be more than 0. Not sufficient. (2) The ratio of the number of red chips to the number of green chips is 4:1. The same info as above. Not sufficient. (1)+(2) Both statements tell the same thing, so we have no new info. Not sufficient. Answer: E
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Re: M2527
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30 Oct 2014, 05:48
The question seems tricky, but its one of those that you don't need scratch paper for.
The main thing to recognize is that both statements say the same thing... so if either is sufficient both are and vice versa. In this case, the prompt asks for the probability of picking two greens without replacement. Both statements say that we have red to green chips in a 4:1 ratio. However, we don't' know whether we have 5 total chips (prob = 0 with only 1 green chip) or any higher mutliple of 5 (prob > 0).
E



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Re: M2527
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23 Nov 2014, 16:06
Hi Bunuel, One more comment... I think that even if the question were to tell us that are indeed more than 1 green chip, the answer will still be E. Example 1: 8 red chips and 2 green chipsProbability of both green = 2/8 * 1/7 = 1/28 Example 2: 12 red chips and 3 green chipsProbability of both green = 3/12 * 2/11 = 1/22 Example 3: 16 red chips and 4 green chipsProbability of both green = 4/16 * 3/15 = 1/20 Can we then conclude that the more quantity overall, the more chances you have of getting a green chip?
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25 Nov 2014, 06:19
minwoswoh wrote: Hi Bunuel,
One more comment... I think that even if the question were to tell us that are indeed more than 1 green chip, the answer will still be E.
Example 1: 8 red chips and 2 green chips Probability of both green = 2/8 * 1/7 = 1/28
Example 2: 12 red chips and 3 green chips Probability of both green = 3/12 * 2/11 = 1/22
Example 3: 16 red chips and 4 green chips Probability of both green = 4/16 * 3/15 = 1/20
Can we then conclude that the more quantity overall, the more chances you have of getting a green chip? It should be: Example 1: 8 red chips and 2 green chipsProbability of both green = 2/10 * 1/9 = 1/45 Example 2: 12 red chips and 3 green chipsProbability of both green = 3/15 * 2/14 = 1/35 Example 3: 16 red chips and 4 green chipsProbability of both green = 4/20* 3/19 = 3/95 The probability of choosing 1 green chip if the ratio of red to green is 4:1 will be the same (1/5) no matter how many total chips we have. The probability of choosing 2 green chips will increase by increasing the total number of chips. Hope it's clear.
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Re: M2527
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25 Nov 2014, 10:08
Bunuel wrote: minwoswoh wrote: Hi Bunuel,
One more comment... I think that even if the question were to tell us that are indeed more than 1 green chip, the answer will still be E.
Example 1: 8 red chips and 2 green chips Probability of both green = 2/8 * 1/7 = 1/28
Example 2: 12 red chips and 3 green chips Probability of both green = 3/12 * 2/11 = 1/22
Example 3: 16 red chips and 4 green chips Probability of both green = 4/16 * 3/15 = 1/20
Can we then conclude that the more quantity overall, the more chances you have of getting a green chip? It should be: Example 1: 8 red chips and 2 green chipsProbability of both green = 2/10 * 1/9 = 1/45 Example 2: 12 red chips and 3 green chipsProbability of both green = 3/15 * 2/14 = 1/35 Example 3: 16 red chips and 4 green chipsProbability of both green = 4/20* 3/19 = 3/95 The probability of choosing 1 green chip if the ratio of red to green is 4:1 will be the same (1/5) no matter how many total chips we have. The probability of choosing 2 green chips will increase by increasing the total number of chips. Hope it's clear. Such a careless mistake from my part! It´s crystal clear now, Bunuel. This was very helpful for me to understand how DS tests the concept of Ratios vs. Concrete Numbers in Probability (since obviously, probability is expressed as a ratio itself...)
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Re: M2527
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18 Jan 2015, 04:13
I think this question is good and helpful.



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31 Aug 2017, 21:53
Good question and Insight full.



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I selected D because i dint read that the chips are to be drawn without replacement. I would just like to add my two cents to the problem. If the question were with replacement, then the probablity of picking a green chip everytime (irrespective of the number of times i pick it) will be 1/5 or 20/100. In this case both the statements are sufficient alone and the ratio of quantity is sufficient to answer the question. But since this is a gmatclub quant test (expect traps & pitfalls ) and the questions states explicitly that the the second draw is made without replacement, there is no way to figure out the probablity of choosing green chip in second selection. We would need the exact number of chips to figure out the probability!  Crave Your Rave 



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Re: M2527
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13 Jun 2018, 03:03
Bunuel wrote: Official Solution:
(1) 20% of all chips in the basket are green. Therefore 80% of all chips in the basket are red, which means that the ratio of the number of red chips to the number of green chips is 4:1 (80:20). Now, if there are total of 5 chips in the basket (4 red + 1 green) then the probability that both chips will be green will be 0 (since there are NOT two chips in a basket) but if there are total of 10 chips in the basket (8 red + 2 green) then the probability that both chips will be green will be more than 0. Not sufficient. (2) The ratio of the number of red chips to the number of green chips is 4:1. The same info as above. Not sufficient. (1)+(2) Both statements tell the same thing, so we have no new info. Not sufficient.
Answer: E BunuelHi Bunuel can we solve it this way? total number of chips = r + g no of ways 2 green can be drawn = gC2 = g(g1) no of ways 2 chips can be drawn from the combined pool =(r+g)C2 = (r+g)(r + g ) probability that both chips will be green = \frac{g(g1)}{(r+g)(r+g1)} per statement 1: g = 20% of (r + g) therefore it comes \frac{g}{r+g}=\frac{1}{5} therefore not sufficient per statement 2: \frac{r}{g} = \frac{4}{1} not sufficient combined same situation not sufficient
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Re: M2527
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25 Jun 2018, 04:58
Hello Bunuel , I have a question, if you could please respond  In this question, we will not be able to get the answer as there are cases in which we will not have enough green balls. Lets say , in all the case we have enough red and green ball  then if just the ratios of ball is given , can we get the probability ?? My doubt is  lets say ratio is 3:2 (r,b) and target is to find 2 balls of r color. if sample space is 5  3c2/5c2 ;  case 1 if sample space is 15  9c2 / 15c2; case 2 answer of case 1 and case 2 will always differ  i think, we can find the probability just with the proportion, only if we need to fetch one object. (obviously with other conditionb links we should have the obj available in all the cases). Kindly clarify .



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Re: M2527
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29 Sep 2018, 10:30
Hi Bunuel, In case there is information available that there are more than 1 green chip, you have mentioned that probability of 1st chip will be same but for 2nd one it increases as total no. of chips increases. But total probability for both chips will be different with different number of total chips. So answer would have been 'E', am i right?










