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Math Expert V
Joined: 02 Sep 2009
Posts: 59727

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15 00:00

Difficulty:   55% (hard)

Question Stats: 54% (01:29) correct 46% (01:07) wrong based on 343 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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Math Expert V
Joined: 02 Sep 2009
Posts: 59727

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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.

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Manager  Joined: 12 Sep 2014
Posts: 141
GMAT 1: 740 Q49 V41 GPA: 3.94

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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
Manager  Joined: 10 May 2014
Posts: 135

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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?
Math Expert V
Joined: 02 Sep 2009
Posts: 59727

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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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Manager  Joined: 10 May 2014
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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...)
Intern  Joined: 12 Mar 2014
Posts: 5
Schools: Stanford '20

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I think this question is good and helpful.
Intern  B
Joined: 29 Dec 2016
Posts: 1

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Good question and Insight full.
Intern  B
Joined: 26 May 2017
Posts: 25
GMAT 1: 620 Q48 V27 ### Show Tags

1
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!

Intern  B
Joined: 29 Mar 2019
Posts: 1

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I think this is a high-quality question and I agree with explanation. Bunuel, you are the BEST!
Intern  B
Joined: 24 Feb 2014
Posts: 38
Location: United States (GA)
WE: Information Technology (Computer Software)

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I think this is a high-quality question and I agree with explanation.
Intern  B
Joined: 20 Apr 2019
Posts: 7

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Can this be generalised into a learning that its not possible to calculate probability of an event if only the ratios of the number of objects are given? Re: M25-27   [#permalink] 07 Nov 2019, 02:38
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# M25-27

Moderators: chetan2u, Bunuel  