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GMAT Diagnostic Test Question 37 [#permalink]
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07 Jun 2009, 01:00
GMAT Diagnostic Test Question 37Field: probability Difficulty: 700 A box has 6 red hats and 5 green hats. What is the probability of drawing at least one green hat in two consecutive drawings if the hat is not replaced? A. \(\frac{10}{11}\) B. \(\frac{8}{11}\) C. \(\frac{7}{12}\) D. \(\frac{5}{13}\) E. \(\frac{2}{7}\)
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Re: GMAT Diagnostic Test Question 37 [#permalink]
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13 Jun 2009, 21:08
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ExplanationOfficial Answer: BProbability (at least 1 green hat) = Probability (2 hats)  Probability (no green hat) The probability of favourable outcome equals favourable outcomes divided by total outcomes: \(\frac{C_{11}^2  C_6^2}{C_{11}^2} = \frac{55  15}{55} = \frac{40}{55} = \frac{8}{11}\) So the correct answer is B.
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Re: GMAT Diagnostic Test Question 38 [#permalink]
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28 Jul 2009, 19:50
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GMAT TIGER wrote: Explanation
Official Answer: BProbability (at least 1 green hat) = Probability (2 hats)  Probability (no green hat) The probability of favourable outcome equals favourable outcomes divided by total outcomes: \(\frac{C_{11}^2  C_6^2}{C_{11}^2} = \frac{55  15}{55} = \frac{40}{55} = \frac{8}{11}\) So the correct answer is B. Hi could you please advise on this, there are two consecutive draw, so the probability of drawing atleast 1 green hat = (probability of drawing one green ball + probability of drawing one red ball) + (probability of drawing 2 green balls) which is (5c1/11c1*6c1/10c1 ) + (5c1/11c1 *4c1/10c1) = 50/110 =5/11



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Re: GMAT Diagnostic Test Question 38 [#permalink]
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31 Jul 2009, 04:36
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I use a bit different approach to solve it: Probability of at least 2 green hats = P(RG)+P(GR)+P(GG) = (6/11)*(5/10)+(5/11)*(6/10)+(5/11)*(4/10)=6/22+6/22+4/22=16/22=8/11 to me it was less time consuming.



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Re: GMAT Diagnostic Test Question 38 [#permalink]
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11 Aug 2009, 08:13
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For me easiest way is 1(RR)=16/11*5/10=8/11



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Re: GMAT Diagnostic Test Question 38 [#permalink]
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14 Aug 2009, 20:56
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I think this is the best approach. Especially when there are multiple situations that satisfy the question. In this question, it was simple, there are only 4 possble results RR, RG, GR, GG, and the only one that doesn't satisfy the solution is RR. So rather than finding the probability of RG, GR, and GG that DOES satisfy the "at least 1 green hat". If we find the situations that DO satisfy, meaning all 3 situations, then we have to add together their total probability. If we find the probability of the 1 situation we DO NOT want, all we have to do is subtract that probability from 1 and it's a much simpler problem. I would say much simpler than a 700 level question. hrish88 wrote: For me easiest way is 1(RR)=16/11*5/10=8/11
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Re: GMAT Diagnostic Test Question 38 [#permalink]
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11 Sep 2009, 10:38
Yeah... whenever you see the word "atleast" it's best to apply the reverse approach ... i.e.
P (Atleast one Green) = 1  P(No Green)



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Re: GMAT Diagnostic Test Question 38 [#permalink]
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04 Jan 2010, 22:38
hrish88 wrote: For me easiest way is 1(RR)=16/11*5/10=8/11 I did it the same way. I think this is the best way because it saves time. The math in gmat is all about doing it with the least amount of time.



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Re: GMAT Diagnostic Test Question 38 [#permalink]
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20 Feb 2010, 17:08
wilbase wrote: hrish88 wrote: For me easiest way is 1(RR)=16/11*5/10=8/11 I did it the same way. I think this is the best way because it saves time. The math in gmat is all about doing it with the least amount of time. I agree...time management is crucial at the test and for probability questions P(wanted) = 1  P (not wanted) is very usefull P(at least 1 Green) = 1  P(not one green) = 1  P(all red) = 1  (6/11 * 5/10) = 1  30/110 = 1  3/11 = 8/11



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Re: GMAT Diagnostic Test Question 38 [#permalink]
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07 May 2010, 16:50
The probability of two green balls in succession (going by 1  Probability of 2 red balls) is 8/11 and this makes sense. However, cant we get the calculation of p(1st green ball) = 5/11 and probability of drawing 2nd green ball (p 2nd green ball) = 4/10 so 5/11 * 4/10 = 2/11. What is wrong with this approach?



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Re: GMAT Diagnostic Test Question 38 [#permalink]
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07 May 2010, 17:03
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vsrinivasan wrote: The probability of two green balls in succession (going by 1  Probability of 2 red balls) is 8/11 and this makes sense. However, cant we get the calculation of p(1st green ball) = 5/11 and probability of drawing 2nd green ball (p 2nd green ball) = 4/10 so 5/11 * 4/10 = 2/11. What is wrong with this approach? This is wrong. See you have three cases. RG, GR , GG i.e. one of them is green or both of them is green. P(RG) = P(GR) = 6/11 * 5/10 = 3/11 > You have missed this caseP(GG) = 5/11 * 4/10 = 2/11 > You have considered only this casetotal probability = P(RG) + P(GR) + P(GG) = 3/11 + 3/11 + 2/11 = 8/11
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Re: GMAT Diagnostic Test Question 38 [#permalink]
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12 Sep 2010, 13:47
The way i did was, 3 possibilities, (A) both drawn are green:2/11 (B) 1st green 2nd red :3/11 (C) 1st red 2nd green :3/11. Hence total probability of atleast 1 green=P(A)+P(B)+P(C)=8/11.



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Re: GMAT Diagnostic Test Question 38 [#permalink]
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15 Nov 2010, 22:28
I think perhaps one of the answer choices should be 3/11, a trick to catch someone not paying 100% attention is one I see frequently on other questions. Thanks for the test, really helpful.



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Re: GMAT club test problem (probability) [#permalink]
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26 Jun 2011, 11:48
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6R 5G
Total = 11
no replacements
Probability of selecting atleast one green = 1 probability of selecting no green = 1  probability of selecting red = 1  (6/11)(5/10) =8/11
Answer is B.



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Re: GMAT Diagnostic Test Question 38 [#permalink]
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01 Nov 2011, 03:58
P(GG)+P(GR)+P(RG) = 5/11*4/10+5/11*6/10+6/11*5/10 = 8/11



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Re: GMAT Diagnostic Test Question 38 [#permalink]
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18 Jul 2012, 14:56
Easiest way to do this type of question is to analize the worst case scenario
If you want the probability of drawing at least one green hat in two consecutive drawings, first count the possible outcomes drawing red hats
1 red hat, 1 rh, 1 rh, 1 rh, 1 rh, 1 rh.
Ok, so now we have exausted all possible outcomes without drawing any green hat. Subsequent to this count (1 red hat, 1 rh, 1 rh, 1 rh, 1 rh, 1 rh), start drawing green hats until you have the desired result, which is at least 2 concecutive green hats. This final count adds up to 8 drawings. Finally, divide 8 drawings by the possible number of hats in total. So, the probability of drawing at least one green hat in two consecutive drawings is 8/11.



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Re: GMAT Diagnostic Test Question 38 [#permalink]
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19 Jul 2012, 02:53
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Re: GMAT Diagnostic Test Question 38 [#permalink]
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21 Jul 2012, 15:13
the cases are:
a. Both drawn are Green: 5/11*4/10 b. First Green and then Red: 5/11*6/10 c. First Red and then Green: 6/11*5/10
Thus, the total probability is: a+b+c = 80/110 = 8/11



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Re: GMAT Diagnostic Test Question 38 [#permalink]
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14 Aug 2013, 18:08
Okay, I get the inverse probability idea, but let's say you do this,
P(GG) + P(GR) + P(RG) = correct answer I know that the correct answer is 8/11, which you can get to like this: P(GG) = 5/11 * 4/10 = 20/110 P(GR) = 5/11 * 6/10 = 30/110 P(RG) = 6/11 * 5/10 = 30/110 so correct answer = 80/110 = 8/11
However, I'm weak on the concept of P(GG) and could use some strengthening there. Why don't you multiply 20/110 by 2 to reflect that GG could be ordered either way?



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Re: GMAT Diagnostic Test Question 38 [#permalink]
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15 Aug 2013, 03:06




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