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GMAT Diagnostic Test Question 37 [#permalink]
 07 Jun 2009, 00:54
GMAT Diagnostic Test Question 37Field: probability Difficulty: 600-650
A jar contains B blue balls, 6B + 10 yellow balls and 2B+5 green balls. If there are only blue, yellow and green balls in the jar, what is the probability of taking out a blue or green ball? A. \frac{1}{5}B. \frac{1}{4}C. \frac{1}{3}D. \frac{1}{2}E. \frac{2}{3}
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Re: GMAT Diagnostic Test Question 37 [#permalink]
13 Jun 2009, 21:46
Explanation
Official Answer: C Total = B + 6B + 10 + 2B + 5 = 9B + 15 Probability of getting a blue or green = \frac{(B + 2B + 5)}{(9B + 15)} = \frac{1}{3}. So the answer is C.
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Re: GMAT Diagnostic Test Question 37 [#permalink]
18 Jul 2009, 05:11
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P(Green or Yellow) = 1 - P(Blue)
ie P(Green or Yellow) = 1 - ( B/(9B+15)
ie P(Green or Yellow) = (8B + 15)/(9B + 15)...
How can it be solved??? Please correct if I am wrong..
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Re: GMAT Diagnostic Test Question 37 [#permalink]
18 Jul 2009, 05:33
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Here again, in the explanation you wrote Quote: Probability (Green or Yellow) = (B + 2B + 5) / (9B + 15) = 1/3. but Green or Yellow would be "( 6B + 10 + 2B + 5)" and that would equal (8B + 15) / (9B + 15), which is what Kyabe got. Hope to be right this time!
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Re: GMAT Diagnostic Test Question 37 [#permalink]
18 Jul 2009, 07:21
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The question stem is incorrect as per your calculation.
I think the question should ask for the probability of taking out a blue or a green ball.
then it would be 1/3.
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Re: GMAT Diagnostic Test Question 37 [#permalink]
18 Jul 2009, 14:35
Thank you guys - I will let GT tackle this one. Please keep the questions and suggestions coming!
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Re: GMAT Diagnostic Test Question 37 [#permalink]
18 Jul 2009, 21:53
Sorry guys for the confusion. It is updated as per your suggestion. Thanks for the suggestion. bb wrote: GMAT Diagnostic Test Question 37Field: probability Difficulty: 600-650
A jar contains B blue balls, 6B + 10 yellow balls and 2B+5 green balls. What is the probability of taking out a blue or green ball if there are no other balls? A. \frac{1}{5}B. \frac{1}{4}C. \frac{1}{3}D. \frac{1}{2}E. \frac{2}{3}
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Re: GMAT Diagnostic Test Question 37 [#permalink]
19 Sep 2009, 10:40
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The question doesn't make any sense. The question states "if there are no other balls?" This means if if all the yellow balls have already been taken out, and what's left in the jar is only blue balls and green balls. This can only mean that the probably of selecting a blue or a green is 100%.
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Re: GMAT Diagnostic Test Question 37 [#permalink]
24 Sep 2009, 18:50
linfongyu wrote: The question doesn't make any sense. The question states "if there are no other balls?" This means if if all the yellow balls have already been taken out, and what's left in the jar is only blue balls and green balls. This can only mean that the probably of selecting a blue or a green is 100%. U r right. I also have same confusion.........
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Re: GMAT Diagnostic Test Question 37 [#permalink]
28 Nov 2009, 00:42
I'm like you guys, I'm a bit confused with the question.
What should be understood by the statement "if there are no other balls" ?
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Re: GMAT Diagnostic Test Question 37 [#permalink]
13 Dec 2009, 13:17
I agree with that the phrase, "if there are no other balls" is very confusing in the question. What does this mean?
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Re: GMAT Diagnostic Test Question 37 [#permalink]
17 Dec 2009, 23:19
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I believe that "no other balls" means that there are only blue, yellow and green balls in the jar. The wording is a bit odd, perhaps an edit in the future? Way to solve- the total number of balls is B + 6B+10 + 2B+5 To get Blue or green, you would not want Yellow. So find the prob of getting yellow and subratct from 1. 6B+10 / 9B+15 = 2/3 for all positive values of B which can be assumed in this case. 2/3 chance of drawing a Yellow, meaning 1/3 chance of drawing a green or blue
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Re: GMAT Diagnostic Test Question 37 [#permalink]
22 Dec 2009, 11:24
It helps to think of it as fractions and plug in
Just plug in a value for B(the number of blue balls)- I chose 2 so that means
there are 2 blue balls, 22 yellow balls (6(2)+10), and 9 green balls (2(2)+5) and 33 total
The answer to the question will be the probability of not selecting a yellow ball
1-22/33= 1/3 since 22/33=2/3 so the answer is D 1/3
I hope this helps!
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Re: GMAT Diagnostic Test Question 37 [#permalink]
23 Dec 2009, 04:03
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Hey guys!
I think the key is "blue or green ball". Or means you have to find the probability for each event to occur and then add the probabilities.
The total no of possible outcomes is B+6B+10+2B+5=9B+15
So P(blue occurs)=B/(9b+15)
P(green occurs)=2B+15/9B+15
Add the two fractions and you get 3b+5/3(3B+5)=1/3
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Re: GMAT Diagnostic Test Question 37 [#permalink]
23 Dec 2009, 05:51
How would you guys change the wording to avoid the confusion? I suggest this: Quote: A jar contains B blue balls, 6B + 10 yellow balls and 2B+5 green balls. If there are only blue, yellow and green balls in the jar, what is the probability of taking out a blue or green ball? marcos4 wrote: I'm like you guys, I'm a bit confused with the question.
What should be understood by the statement "if there are no other balls" ?
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Re: GMAT Diagnostic Test Question 37 [#permalink]
23 Dec 2009, 13:50
dzyubam wrote: How would you guys change the wording to avoid the confusion? I suggest this: Quote: A jar contains B blue balls, 6B + 10 yellow balls and 2B+5 green balls. If there are only blue, yellow and green balls in the jar, what is the probability of taking out a blue or green ball? Yup, that works! As it is in sentence correction, ambiguity is evil
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Re: GMAT Diagnostic Test Question 37 [#permalink]
05 Feb 2010, 16:25
dzyubam wrote: How would you guys change the wording to avoid the confusion? I suggest this: Quote: A jar contains B blue balls, 6B + 10 yellow balls and 2B+5 green balls. If there are only blue, yellow and green balls in the jar, what is the probability of taking out a blue or green ball? I was thrown off by the "if there are no other balls" part too. This is much better. Thanks for the super test!
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Re: GMAT Diagnostic Test Question 37 [#permalink]
05 Feb 2010, 18:39
I was also confused by the wording on this one. I like the suggestion above recommending a more precisely worded stem.
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Re: GMAT Diagnostic Test Question 37 [#permalink]
08 Feb 2010, 08:57
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Re: GMAT Diagnostic Test Question 37 [#permalink]
22 May 2010, 08:52
Easy to do . Because B is constant, it does not change, so the number of balls does not change as well. lets B=1 we have the following balls: blue -1 yellow - 16 green - 7total -24 BLUE OR GREEN , MEANS NOT YELLOW , so 1-P(yellow)= 1-16/24=1-2/3=1/3 
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Re: GMAT Diagnostic Test Question 37
[#permalink]
22 May 2010, 08:52
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