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Intern  Affiliations: AIESEC
Joined: 13 Feb 2011
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Smit: K
A box contains 100 balls, numbered from 1 to 100. If three b  [#permalink]

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125 00:00

Difficulty:   65% (hard)

Question Stats: 59% (02:00) correct 41% (02:20) wrong based on 1130 sessions

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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

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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31
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!

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.

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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2

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.
##### General Discussion
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mbafall2011 wrote:

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.

OA is 1/2.

It doesn't matter for order whether it's with or without replacement case. EEO, EOE, and OEE are 3 different scenarios and each has the probability of 1/8, so the probability of two even numbered balls and one odd numbered ball is 3*1/8.

Total: 1/4+3/8=1/2.
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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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Posts: 65001
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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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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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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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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
OOO

Even sum:
EEE
EOO
OEO
OOE

Now, 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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Bunuel - u r amazing.great explanation from all aspects. thanks. +1
+1 fluke
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P(OOO)=(50/100)^3 = (1/2)^3= 1/8
P(EEO)=3*(50/100)^2*50/100 =3*(1/2)^2*1/2=3/8

Finally= 1/8+3/8=1/2 Ans.
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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 ?

You have to purchase them from mba.com,

However, if you're looking for Combinatorics/ probability questions, you can find them here: permutations-combinations-probability-download-questions-57156.html
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Re: A box contains 100 balls, numbered from 1 to 100. If three b  [#permalink]

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Since we have equal number of odd and even numbers (with replacement) isn't it self-explanatory 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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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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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

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.
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Bunuel 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 3

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
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garazhaka wrote:
Bunuel 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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Re: A box contains 100 balls, numbered from 1 to 100. If three b  [#permalink]

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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  [#permalink]

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1
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.

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Re: A box contains 100 balls, numbered from 1 to 100. If three b  [#permalink]

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I reasoned this out the following way, is it correct? Regardless of the order we have 4 total scenarios (combinations) of even-odd 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 2-2-1 or 2-1-2) we will get an odd number in 2 combinations out 4 possible.

Hence 1/2 Re: A box contains 100 balls, numbered from 1 to 100. If three b   [#permalink] 16 Nov 2015, 09:41

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