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# M28-45

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16 Sep 2014, 01:31
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95% (hard)

Question Stats:

46% (03:53) correct 54% (01:53) wrong based on 114 sessions

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Two marbles are drawn from a jar with 10 marbles. If all marbles are either red or blue, is the probability that both marbles selected will be red greater than $$\frac{3}{5}$$?

(1) The probability that both marbles selected will be blue is less than $$\frac{1}{10}$$.

(2) At least 60% of the marbles in the jar are red.

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16 Sep 2014, 01:31
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Official Solution:

The question asks whethewr $$P(R \ and \ R)=\frac{R}{10}*\frac{R-1}{9} \gt \frac{3}{5}$$?

Is $$R(R-1) \gt 54$$?

Is $$R \gt 7$$? (By number plugging) So, the question asks whether the number of red marbles is more than 7 (8, 9, or 10).

(1) The probability that both marbles selected will be blue is less than $$\frac{1}{10}$$. This implies that $$\frac{B}{10}*\frac{B-1}{9} \lt \frac{1}{10}$$. So, we have that $$B(B-1) \lt 9$$, thus $$B \lt 4$$, so the number of red marbles in the jar is 7, 8, 9, or 10. Not sufficient.

(2) At least 60% of the marbles in the jar are red. This implies that the number of red marbles is 6 or more. Not sufficient.

(1)+(2) From above we have that $$R \gt 6$$. Not sufficient.

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07 Jun 2015, 08:25
Bunnel

How r(r-1)>54
is 7?

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07 Jun 2015, 09:31
shrutikasat wrote:
Bunnel

How r(r-1)>54
is 7?

Bu number plugging. R is an integer, if R = 7, then R(R - 1) = 42 < 54 but if R = 8, then R(R - 1) = 56 > 54, thus R must be greater than 7.
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08 Jun 2015, 08:33
we have 10 marbles blue and red.. we need to check the probability of selecting both red marbles f is greater than 3/5 i.e 0.6
stmt 1 - we have probability of both blue is < 1/10
so b/10 * (b-1)/9 < 1/10
so b(b-1) < 90/10
b(b-1) < 9
so b can be 2,3
if b is 2 then the probability of both red marbles is 8/10 * 7/9 which is 0.622 hence greater than 3/5
but if b is 3 then the probability of both red marbles is 7/10 * 6/9 which is 0.466 so we cant come down to a single result hence not sufficient

stmt 2 - we have 60% of the marbles are red so again we have the same confusion as in stmt 1 when the red marbles are 7 or when they are 8
so not sufficient

combining also we do not come to any conclusion so ans should be E
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30 Jun 2015, 21:17
Hi Bunuel, I don't understand, where am I mistaken?

If we have 6 red balls and 4 blue balls the probability of picking 2 reds in a row is:

6/10 * 5/9 = 30/90 simplified to 1/3, bigger than 3/5. Why is 6 red balls not sufficient?

Thanks.
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01 Jul 2015, 01:57
Bombardeiro wrote:
Hi Bunuel, I don't understand, where am I mistaken?

If we have 6 red balls and 4 blue balls the probability of picking 2 reds in a row is:

6/10 * 5/9 = 30/90 simplified to 1/3, bigger than 3/5. Why is 6 red balls not sufficient?

Thanks.

How is 1/3 = 0.333... greater than 3/5 = 0.6 ???
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01 Jul 2015, 20:09
1
Bunuel wrote:
Two marbles are drawn from a jar with 10 marbles. If all marbles are either red or blue, is the probability that both marbles selected will be red greater than $$\frac{3}{5}$$?

(1) The probability that both marbles selected will be blue is less than $$\frac{1}{10}$$.

(2) At least 60% of the marbles in the jar are red.

------
Nice meaty problem and Bunuel has explained. Let me try and explain in a different manner.

We are trying to know whether there are more than 7 red balls, that is 8 or 9 red balls out of the 10 since both of these will make the possibility of two consecutive reds > 3/5. (7 reds will give us 7/10*6/9=14/30 which is < 3/5...so we need at least 8).

1. Probability of two consecutive blues is less than 1/10th. For this to be true, there must be either 1 blue, 2 blues, or 3 blues. Based on what I noted earlier, 1 and 2 blues will answer the target question as yes but 3 blues will make the answer no. Insufficient.

2. Tells us that there can be 7,8, or 9 red balls. We need to know if there are 8 or greater to answer definitely. Insufficient.

Combined: We can still have 1 blue and 9 reds in which case target answer is yes or 3 blues and 7 reds in which case target answer is no.
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26 Jul 2015, 00:13
I think this is a high-quality question and I agree with explanation. When it says all marbles are either red or blue, does that mean that there is at least one of each in each in the total of 10? If yes, then in the explanation of the first statement, Red can not be 10. Maximum could be 7,8,9. Please clarify if you can. Thanks,

sal
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10 Jan 2016, 18:55
I did it similarly to Aviram above:

Question: Is the chance of two reds greater than .6?

So since there are only ten numbers and the chance of two reds is fairly high , I plugged in starting with 7.

(7/10) * (6/9) = (7*3*2/2*5*3*3) = (7/15) which is about .46

Then I tried plugging in 8 which is 56/90 = .62 which is very close so the red marbles have to be either 8, 9, or 10.

Statement 1: LESS than 1/10 could mean 0 which would make the statement "YES". So find the largest number possible to make it "no" and plug in:

I tried 4 first:
(4/10)* (3/9) = .13 very close to what I need but a bit too high so blue marbles = 3 making the statement "NO" so it's insufficient.

Statement 2: This is tricky because the word AT LEAST means there could be 6 OR greater red marbles making the statement insufficient.
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14 Oct 2016, 01:30
I think this is a poor-quality question and I don't agree with the explanation. Two marbles are drawn from a jar with 10 marbles. If all marbles are either red or blue, is the probability that both marbles selected will be red greater than..?
In this question it should be mentioned whether the balls are drawn simultaneously or one after the other without replacement (which is the case). I assumed that 2 balls are drawn simultaneously and then I got confused between B and D
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14 Oct 2016, 01:53
gmatdemolisher1234 wrote:
I think this is a poor-quality question and I don't agree with the explanation. Two marbles are drawn from a jar with 10 marbles. If all marbles are either red or blue, is the probability that both marbles selected will be red greater than..?
In this question it should be mentioned whether the balls are drawn simultaneously or one after the other without replacement (which is the case). I assumed that 2 balls are drawn simultaneously and then I got confused between B and D

No, that's not correct.

1. If the drawing is with replacement it's explicitly mentioned. If it's not mentioned, then it;s without replacement.

2. Mathematically the probability of picking two balls simultaneously, or picking them one at a time (without replacement) is the same.
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17 Aug 2017, 23:58
I think this is a high-quality question and I agree with explanation.
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26 Aug 2017, 10:43
I think this is a poor-quality question and I don't agree with the explanation. The solution seems to be wrong. The statement one clearly gives the probability of both the marbles being blue. Since the jar contains only blue and red marbles it is understood that if the probability of picking two blue marbles is less than 1/10 then the probability of picking two red marbles will be greater than 9/10.
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27 Aug 2017, 04:01
abhishekvigg wrote:
I think this is a poor-quality question and I don't agree with the explanation. The solution seems to be wrong. The statement one clearly gives the probability of both the marbles being blue. Since the jar contains only blue and red marbles it is understood that if the probability of picking two blue marbles is less than 1/10 then the probability of picking two red marbles will be greater than 9/10.

This is wrong. The probability of two marbles being blue and the probability of two marbles being red are not complementary. So, P(RR) + P(BB) ≠ 1. It's P(RR) + P(BB) + P(BR) = 1. The solution above shows cases which give two different answers to the question.

If R = 7 and B = 3 (notice that both statements are satisfied), then the answer is NO.
If R = 8 and B = 2 (notice that both statements are satisfied), then the answer is NO.
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22 Sep 2017, 22:37
I think this is a high-quality question and I agree with explanation. // good question // Approach is worth noting and understanding for future use //
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17 Oct 2017, 20:18
I think this is a high-quality question and I agree with explanation. Very good question. Very clear and simple explanation.
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23 Mar 2018, 22:46
Bunuel wrote:
gmatdemolisher1234 wrote:
I think this is a poor-quality question and I don't agree with the explanation. Two marbles are drawn from a jar with 10 marbles. If all marbles are either red or blue, is the probability that both marbles selected will be red greater than..?
In this question it should be mentioned whether the balls are drawn simultaneously or one after the other without replacement (which is the case). I assumed that 2 balls are drawn simultaneously and then I got confused between B and D

No, that's not correct.

1. If the drawing is with replacement it's explicitly mentioned. If it's not mentioned, then it;s without replacement.

2. Mathematically the probability of picking two balls simultaneously, or picking them one at a time (without replacement) is the same.

Hi Bunuel,

While I do agree on the first point, for the second point, I have my doubts. How are simultaneously and picking one after another the same thing? Suppose we have x red balls out of ten balls, then picking two simultaneously gives us (xC2/10C2) = x(x-1)/90 and if picked one after the other gives us x(x-1)/45. As you see both expressions are different. Can you please help?
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24 Mar 2018, 00:55
SajjitaKundu wrote:
Bunuel wrote:
gmatdemolisher1234 wrote:
I think this is a poor-quality question and I don't agree with the explanation. Two marbles are drawn from a jar with 10 marbles. If all marbles are either red or blue, is the probability that both marbles selected will be red greater than..?
In this question it should be mentioned whether the balls are drawn simultaneously or one after the other without replacement (which is the case). I assumed that 2 balls are drawn simultaneously and then I got confused between B and D

No, that's not correct.

1. If the drawing is with replacement it's explicitly mentioned. If it's not mentioned, then it;s without replacement.

2. Mathematically the probability of picking two balls simultaneously, or picking them one at a time (without replacement) is the same.

Hi Bunuel,

While I do agree on the first point, for the second point, I have my doubts. How are simultaneously and picking one after another the same thing? Suppose we have x red balls out of ten balls, then picking two simultaneously gives us (xC2/10C2) = x(x-1)/90 and if picked one after the other gives us x(x-1)/45. As you see both expressions are different. Can you please help?

$$\frac{xC2}{10C2} = \frac{x!}{(x-2)!2!}*\frac{8!2!}{10!}=\frac{(x-1)x}{90}$$.

$$\frac{x}{10}*\frac{(x-1)}{9}=\frac{(x-1)x}{90}$$.
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27 Mar 2018, 02:21
Option A: I agree with how you derived at B*(B-1) <9
After which it is said that B<4.
Shouldn't it be B<3.

By plugging in : 3*2 =6 < 9
4*3 =12 not <9

Where am I going wrong?
Re: M28-45 &nbs [#permalink] 27 Mar 2018, 02:21

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# M28-45

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