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15 Sep 2014, 23:13



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Re: D0136
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10 Sep 2015, 07:30
Bunuel wrote: Official Solution:
A jar contains only \(B\) blue balls, \(6B + 10\) yellow balls and \(2B+5\) green balls. If one ball is picked at random from the jar what is the probability of getting a blue or green ball?
A. \(\frac{1}{5}\) B. \(\frac{1}{4}\) C. \(\frac{1}{3}\) D. \(\frac{1}{2}\) E. \(\frac{2}{3}\)
Total # of balls in the jar is \(B + (6B + 10) + (2B + 5) = 9B + 15\); The probability of picking a blue or green ball is \(\frac{B+(2B + 5)}{9B + 15}=\frac{1}{3}\).
Answer: C Hey Bunuel, Is the trick to simplifying that expression into 1/3 just simply seeing that the ratio of top to bottom is 1:3? Sometimes I get caught up doing a problem and try to simplify algebraically without "zooming out" for a second to see that. Would you mind showing how you would simplify that algebraically just to satisfy my curiosity?



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11 Sep 2015, 00:48
bigchuck wrote: Bunuel wrote: Official Solution:
A jar contains only \(B\) blue balls, \(6B + 10\) yellow balls and \(2B+5\) green balls. If one ball is picked at random from the jar what is the probability of getting a blue or green ball?
A. \(\frac{1}{5}\) B. \(\frac{1}{4}\) C. \(\frac{1}{3}\) D. \(\frac{1}{2}\) E. \(\frac{2}{3}\)
Total # of balls in the jar is \(B + (6B + 10) + (2B + 5) = 9B + 15\); The probability of picking a blue or green ball is \(\frac{B+(2B + 5)}{9B + 15}=\frac{1}{3}\).
Answer: C Hey Bunuel, Is the trick to simplifying that expression into 1/3 just simply seeing that the ratio of top to bottom is 1:3? Sometimes I get caught up doing a problem and try to simplify algebraically without "zooming out" for a second to see that. Would you mind showing how you would simplify that algebraically just to satisfy my curiosity? \(\frac{B+(2B + 5)}{9B + 15}=\frac{3B+5}{3*(3B+5)}=\frac{1}{3}\).
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Re: D0136
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31 Oct 2015, 20:44
I doubt this is a 700 lvl question...Even I, not a lover or probability questions, solved it under 1 minute: B+2b+5 / 9B+15 3B+5 / 3(3B+5) => simplify by 3B+5 and get 1/3



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Re: D0136
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21 Feb 2016, 09:47
Hi Is it because only one ball is being selected that we are not applying the intersection part where both are possible?? Bunuel wrote: Official Solution:
A jar contains only \(B\) blue balls, \(6B + 10\) yellow balls and \(2B+5\) green balls. If one ball is picked at random from the jar what is the probability of getting a blue or green ball?
A. \(\frac{1}{5}\) B. \(\frac{1}{4}\) C. \(\frac{1}{3}\) D. \(\frac{1}{2}\) E. \(\frac{2}{3}\)
Total # of balls in the jar is \(B + (6B + 10) + (2B + 5) = 9B + 15\); The probability of picking a blue or green ball is \(\frac{B+(2B + 5)}{9B + 15}=\frac{1}{3}\).
Answer: C



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Re: D0136
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21 Feb 2016, 10:10
sinhap07 wrote: Hi Is it because only one ball is being selected that we are not applying the intersection part where both are possible?? Bunuel wrote: Official Solution:
A jar contains only \(B\) blue balls, \(6B + 10\) yellow balls and \(2B+5\) green balls. If one ball is picked at random from the jar what is the probability of getting a blue or green ball?
A. \(\frac{1}{5}\) B. \(\frac{1}{4}\) C. \(\frac{1}{3}\) D. \(\frac{1}{2}\) E. \(\frac{2}{3}\)
Total # of balls in the jar is \(B + (6B + 10) + (2B + 5) = 9B + 15\); The probability of picking a blue or green ball is \(\frac{B+(2B + 5)}{9B + 15}=\frac{1}{3}\).
Answer: C Yes, we are selecting only one ball and we want it to be either blue or green.
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Re: D0136
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30 Jun 2016, 14:14
hi EMPOWERgmatRichC , can you show how to apply the 'Test It' method to this question? thanks a lot



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09 Jul 2016, 08:28
I think this is a highquality question and I agree with explanation. I still do not understand how we get from 3B+5 / 9B+15 to 1/3 ..
Even if I divide by 3B+5 then I end up with 1/3B+3



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09 Jul 2016, 08:31
I think this is a highquality question and I agree with explanation. Someone explained it to me.....EASY....
factoring out is key.
3B+5/9B+15 = 3B+5/ 3(3B+5)  divide both now by 3B+5 and you are left with 1/3



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09 Jul 2016, 08:31



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09 Jul 2016, 10:04
Yes indeed. I find it easy to forget about the basics when you see something visually confusing. Again I appreciate that some people are really into math and left brained but put yourselves in the shoes of those who are right brained. It's like someone who speaks with someone they like and suddenly forget how to speak because they are nervous ! Posted from my mobile device



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10 Jul 2016, 03:37
whitehalo wrote: hi EMPOWERgmatRichC , can you show how to apply the 'Test It' method to this question? thanks a lot Hi, As Probability is ratio between two things so we can plugin number to make it easy. Put B=1 We have 1 blue, 16 Yellow, 7 green, total balls= 24 We need prob. of choosing blue or green Prob of choosing blue = 1/24 Prob of choosing green = 7/24 Prob of choosing Blue ORGreen= 1/24 +7/24 = 8/24 =1/3 Answer is C I hope it helps



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Re: D0136
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10 Jul 2016, 04:20
And also worth pointing out is the way the question is phrased:
Either  OR, which means we need to "add" the two individual probabilities.
If the question had said: AND I suppose I would have to use both individual probabilities and multiply them instead.



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30 Jun 2017, 22:12
I liked this question. It is a clear example of using "OR" condition.
"OR" means addition.
Blue balls = B/(B+6B+10+2B+5) = B/ (9B+15)
Green balls = (2B+5)/(9B+15)
Total= (B+2B+15)/(9B+15) due to common denominator.
Solve and you will be left with 1/3.
Long process? Yes. But one thing I learnt from the members in GMAT club is don't rely on formulas. Step it out and understand.



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01 Nov 2017, 04:52
I had a doubt regarding the answer. We need to find the probability of getting a blue or a green ball. So 1st we could pick up green then blue or 1st we could pick up blue then green. so shouldn't 1/3 be multiplied by 2. If we were given that 1st green and second is blue then the answer 1/3 would be correct. Looking forward to hearing from you.
Regards



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01 Nov 2017, 04:55










