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M25-27

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Bunuel
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M25-27 [#permalink] New post   16 Sep 2014, 01:23
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55% (01:25) correct 45% (01:07) wrong based on 494 sessions
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A basket contains only red and green chips. If two chips are drawn from the basket at random without replacement, what is the probability that both chips will be green?


(1) 20% of all chips in the basket are green.

(2) The ratio of the number of red chips to the number of green chips is 4:1.
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Bunuel
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Re M25-27 [#permalink] New post   16 Sep 2014, 01:23
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A basket contains only red and green chips. If two chips are drawn from the basket at random without replacement, what is the probability that both chips will be green?

(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 a 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 is only one green chip in the basket). However, if there are a total of 10 chips in the basket (8 red + 2 green), then the probability that both chips will be green will be greater than 0. This statement is not sufficient.

(2) The ratio of the number of red chips to the number of green chips is 4:1.

This information is the same as the information provided in statement (1). Thus, this statement is not sufficient.

(1) + (2) Both statements convey the same information, so we have no new information when combining them. Therefore, the combination of the statements is not sufficient.


Answer: E
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intheend14
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Re: M25-27 [#permalink] New post   30 Oct 2014, 06:48
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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: M25-27 [#permalink] New post   23 Nov 2014, 17:06
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 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?
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Re: M25-27 [#permalink] New post   25 Nov 2014, 07:19
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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 chips
Probability of both green = 2/10 * 1/9 = 1/45

Example 2: 12 red chips and 3 green chips
Probability of both green = 3/15 * 2/14 = 1/35

Example 3: 16 red chips and 4 green chips
Probability 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: M25-27 [#permalink] New post   25 Nov 2014, 11:08
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 chips
Probability of both green = 2/10 * 1/9 = 1/45

Example 2: 12 red chips and 3 green chips
Probability of both green = 3/15 * 2/14 = 1/35

Example 3: 16 red chips and 4 green chips
Probability 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: M25-27 [#permalink] New post   18 Jan 2015, 05:13
I think this question is good and helpful.
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Re: M25-27 [#permalink] New post   31 Aug 2017, 22:53
Good question and Insight full.
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Re: M25-27 [#permalink] New post   15 Sep 2017, 03:54
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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 8-) :cry: ) 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: M25-27 [#permalink] New post   18 Aug 2019, 12:38
I think this is a high-quality question and I agree with explanation. Bunuel, you are the BEST!
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Re: M25-27 [#permalink] New post   22 Sep 2019, 18:31
I think this is a high-quality question and I agree with explanation.
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Re: M25-27 [#permalink] New post   19 Feb 2020, 02:57
Hello Team,

Can the question be solved with the below understanding?

1) let the total no. of chips in the bag = x
no. of green chips = 0.20x

Therefore, if both the chips are green ,
then the prob will be

0.20x/x * 0.19x/x-1

Therefore, this statement is not sufficient.

However if it had been with replacement, then the prob would have been
0.20x/x* 0.20x/ x = This would be sufficient then

Please correct my understanding
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Re: M25-27 [#permalink] New post   19 Feb 2020, 03:03
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nikitamaheshwari wrote:
Hello Team,

Can the question be solved with the below understanding?

1) let the total no. of chips in the bag = x
no. of green chips = 0.20x

Therefore, if both the chips are green ,
then the prob will be

0.20x/x * 0.19x/x-1

Therefore, this statement is not sufficient.

However if it had been with replacement, then the prob would have been
0.20x/x* 0.20x/ x = This would be sufficient then

Please correct my understanding


The red part is not correct.

The probability that both chips will be green without replacement P(GG) = 0.2x/x*(0.2x - 1)/(x - 1).

The probability that both chips will be green wit replacement P(GG) = 0.2x/x*0.2x/x = 1/5*1/5.
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Re: M25-27 [#permalink] New post   14 Jul 2021, 00:56
It tests the concept of Ratios vs. Concrete Numbers in Probability. Got confused.
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Re: M25-27 [#permalink] New post   19 Jul 2021, 19:44
Thanks Bunuel - this is a very good question.

Can I conclude that if the question was about picking just 1 chip, in that case D should have been the answer as ratios will work perfectly.
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Re: M25-27 [#permalink] New post   20 Jul 2021, 01:01
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nagarvikas11 wrote:
Thanks Bunuel - this is a very good question.

Can I conclude that if the question was about picking just 1 chip, in that case D should have been the answer as ratios will work perfectly.


Yes, in this case the answer would be D because from each statement we'd get that the probability of picking a green chip as 1/5.
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Re: M25-27 [#permalink] New post   14 Apr 2023, 08:30
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re M25-27 [#permalink] New post   09 Aug 2023, 10:15
I think this is a high-quality question and I agree with explanation.
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Re M25-27 [#permalink]
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