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A box contains 100 balls, numbered from 1 to 100. If three b
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Updated on: 28 May 2013, 14:54
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A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers on the balls selected from the box will be odd? A. 1/4 B. 3/8 C. 1/2 D. 5/8 E. 3/4
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Originally posted by SmitKhurana on 13 Feb 2011, 10:46.
Last edited by Bunuel on 28 May 2013, 14:54, edited 2 times in total.
Edited the question and added the OA




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13 Feb 2011, 11:14
SmitKhurana wrote: Hello there GMAT enthusiasts!
Surely this finds everyone in great guns towards achieving a perfect GMAT Score, in between came across this very peculiar and relatively difficult question for resolution :
Q. A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the 3 numbers on the balls selected from the box will be odd ? Welcome to GMAT Club! Please read and follow: howtoimprovetheforumsearchfunctionforothers99451.html So please: Provide answer choices for PS questions.Original question is: A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers on the balls selected from the box will be odd?A. 1/4 B. 3/8 C. 1/2 D. 5/8 E. 3/4 The sum of the three numbers on the balls selected from the box to be odd one should select either three odd numbered balls (Odd+Odd+Odd=Odd) or two even numbered balls and one odd numbered ball (Even+Even+Odd=Odd); P(OOO)=(1/2)^3; P(EEO)=3*(1/2)^2*1/2=3/8 (you should multiply by 3 as the scenario of two even numbered balls and one odd numbered ball can occur in 3 different ways: EEO, EOE, or OEE); So finally P=1/8+3/8=1/2. Answer: C. Alternately you can notice that since there are equal chances to get even or odd sum after two selections (for even sum it's EE or OO and for odd sum it's EO or OE) then there is 1/2 chances the third ball to make the sum odd.
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Re: Question from the Official GMAC's GMAT Prep Question Bank
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14 Feb 2011, 11:47
What is the official answer.
Since replacement is involved, i would think the order of the EEO ball being picked does not matter.
Thus P(E)&P(E)&P(O) should be 1/8
1/8+1/8 = 1/4.



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14 Feb 2011, 11:56



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14 Feb 2011, 12:26
The order will matter if there is no replacement:
If the first pick is even, the probability of a second even will be 49/99 and odd will be 50/99.
Also, im looking at these as mutually independent events rather than Probability of EEO +EOE etc.
But if i write out all the possibilities
ooo ooe oeo oee eoo eoe eeo eee
then i can see that 4 out of 8 picks are favorable.
This one is tricky!



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14 Feb 2011, 12:31
mbafall2011 wrote: The order will matter if there is no replacement:
If the first pick is even, the probability of a second even will be 49/99 and odd will be 50/99.
Also, im looking at these as mutually independent events rather than Probability of EEO +EOE etc.
But if i write out all the possibilities
ooo ooe oeo oee eoo eoe eeo eee
then i can see that 4 out of 8 picks are favorable.
This one is tricky! Again order does matter. P(odd sum)=P(EEO)+P(EOE)+P(OEE)+P(OOO)=1/8+1/8+1/8+1/8=1/2. Next, the way you are doing (the red part) is correct only for the cases in which there are equal # of even and odd numbers (for example if there were balls numbered from 1 to 99 this approach wouldn't be corrorect, so after all the probability approach is better).
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Re: Question from the Official GMAC's GMAT Prep Question Bank
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15 Feb 2011, 23:43
Bunuel  can you please do this one If it was without replacement? so ill be sure i understood it the right way? thanks.
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16 Feb 2011, 00:09
144144 wrote: Bunuel  can you please do this one If it was without replacement?
so ill be sure i understood it the right way?
thanks. Without replacement; the condition for getting odd doesn't change; only the probability of picking up the ball does; OEE EOE EEO OOO 50/100*50/99*49/98+50/100*50/99*49/98+50/100*49/99*50/98+50/100*49/99*48/98 =1/2*50/99*1/2+1/2*50/99*1/2+1/2*49/99*25/49+1/2*49/99*24/49 rest can be simplified. Correct me if I am wrong, Bunuel.
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16 Feb 2011, 02:48
fluke wrote: 144144 wrote: Bunuel  can you please do this one If it was without replacement?
so ill be sure i understood it the right way?
thanks. Without replacement; the condition for getting odd doesn't change; only the probability of picking up the ball does; OEE EOE EEO OOO 50/100*50/99*49/98+50/100*50/99*49/98+50/100*49/99*50/98+50/100*49/99*48/98 =1/2*50/99*1/2+1/2*50/99*1/2+1/2*49/99*25/49+1/2*49/99*24/49 rest can be simplified. Correct me if I am wrong, Bunuel. Odd sum: OEE EOE EEO OOOEven sum: EEE EOO OEO OOENow, no matter whether we have with or without replacement case, the probability of red events and the probability of blue events will be symmetrical and equal (because there are equal number of even and odd balls) and since the above events describe all possible outcomes when we pick 3 balls and are mutually exclusive then their sum must be 1: \(P(red)=P(blue)=\frac{1}{2}\). To demonstrate for without replacement case: \(P=3*\frac{50}{100}*\frac{50}{99}*\frac{49}{98}+\frac{50}{100}*\frac{49}{99}*\frac{48}{98}=\frac{3*50*49}{100*99*98}(50+16)=\frac{1}{2*33*2}*66=\frac{1}{2}\). Combinatorial approach for without replacement case: \(P=\frac{C^1_{50}*C^2_{50}+C^3_{50}}{C^3_{100}}=\frac{1}{2}\). Hope it's clear.
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16 Feb 2011, 03:40
Bunuel  u r amazing.great explanation from all aspects. thanks. +1 +1 fluke
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17 Feb 2011, 10:55
Baten80 wrote: SmitKhurana wrote: Hello there GMAT enthusiasts!
Surely this finds everyone in great guns towards achieving a perfect GMAT Score, in between came across this very peculiar and relatively difficult question for resolution :
Q. A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the 3 numbers on the balls selected from the box will be odd ? Please give the link from which i can download the questionsYou have to purchase them from mba.com, However, if you're looking for Combinatorics/ probability questions, you can find them here: permutationscombinationsprobabilitydownloadquestions57156.html
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Re: A box contains 100 balls, numbered from 1 to 100. If three b
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28 Mar 2012, 05:10
Can anyone answer my question? Since we have equal number of odd and even numbers (with replacement) isn't it selfexplanatory that the probability of the sum to be odd will be the same of that to be even = 1/2?? I thing that this approach can be applied at any case with replacement i.e. if we pick 4 or 5 or 6 or 50 etc balls the probability of their sum to be odd (even) will be 1/2. Because in this way the answer can be given in 10 seconds...



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28 Apr 2012, 05:57
Bunuel wrote: Again order does matter. P(odd sum)=P(EEO)+P(EOE)+P(OEE)+P(OOO)=1/8+1/8+1/8+1/8=1/2. Excuse me, but I didn't understand why order does matter? At the end we are looking for the sum of the selected balls and not for the order of the selection, so whether it is 2+2+1 or 1+2+2 or 2+2+1 they are all the same! They all equal 4 which is one possible outcome and not 3



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27 Oct 2013, 23:02
Bunuel wrote: Mochad wrote: Bunuel wrote: Again order does matter. P(odd sum)=P(EEO)+P(EOE)+P(OEE)+P(OOO)=1/8+1/8+1/8+1/8=1/2. Excuse me, but I didn't understand why order does matter? At the end we are looking for the sum of the selected balls and not for the order of the selection, so whether it is 2+2+1 or 1+2+2 or 2+2+1 they are all the same! They all equal 4 which is one possible outcome and not 3Consider below two scenarios: First=Even, Second=Even, Third=Odd; First=Even, Second=Odd, Third=Even; Are these scenarios the same? No. That's why the order matters. Argh... it depends on how you look at the problem. If you calculate your full set of events where the order matters, then the order matters also for the "favorable" set of events. I treated the question where order doesn't matter (because it doesn't matter for summation and because we are allowed to disregard it since the balls are replaceable) and only looked at the end result of the number of balls I had after the selection process was over: 3x Odds 2x Odds + 1x Even 2x Evens + 1x Odd 3x Evens 2 of those are "favorable" (first and third), thus 2/4 = 1/2



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27 Oct 2013, 23:16
garazhaka wrote: Bunuel wrote: Mochad wrote: Consider below two scenarios: First=Even, Second=Even, Third=Odd; First=Even, Second=Odd, Third=Even;
Are these scenarios the same? No. That's why the order matters.
Argh... it depends on how you look at the problem. If you calculate your full set of events where the order matters, then the order matters also for the "favorable" set of events. I treated the question where order doesn't matter (because it doesn't matter for summation and because we are allowed to disregard it since the balls are replaceable) and only looked at the end result of the number of balls I had after the selection process was over: 3x Odds 2x Odds + 1x Even 2x Evens + 1x Odd 3x Evens 2 of those are "favorable" (first and third), thus 2/4 = 1/2 You get the probability of 1/2 in either case. But in this problem the order does matter. For example, the case of EEO is different from EOE.
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16 Aug 2014, 02:26
E = Even, O = Odd EEO = 1st ball is even, 2nd ball is even, 3rd ball is odd Prob(EEO) + Prob(EOE) + Prob(OEE) + Prob(OOO) = 4*Prob(EEO) = 4 * Prob(E)*Prob(E)*Prob(O) = 4*(1/2)^3 = 4/8 = 1/2



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Re: A box contains 100 balls, numbered from 1 to 100. If three b
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05 Sep 2015, 03:13
This is how I did it. Three draws with replacements. So I'll consider each draw as fresh. Now for favorable choices, I need either all 3 as ODD or 1 ODD and 2 EVEN. So for every draw we'll have 50 favorable options. Lets count all favorable options = 50 + 50 + 50 = 150. Total options to pick from = 100 * 3 = 300. Probability = 150/300 = 1/2. +Kudos, if this helped!
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16 Nov 2015, 09:41
I reasoned this out the following way, is it correct? Regardless of the order we have 4 total scenarios (combinations) of evenodd that can impact our game: 1. all even 2. all odd (works for us) 3. 2 even 1 odd (works for us) 4. 2 odd 1 even. The order does not matter because in any order (whether it is 221 or 212) we will get an odd number in 2 combinations out 4 possible. Hence 1/2
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