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100 cubes were placed into a sack and each numbered from 1 [#permalink]
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17 Nov 2007, 16:40
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100 cubes were placed into a sack and each numbered from 1 to 100. If three cubes are to be withdrawn with replacement, What is the probability that the sum of the numbers on the withdrawn 3 cubes will be odd. == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.



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1. 1 cube: odd/even =50%/50% 2. 2 cubes even (odd+odd,even+even)  50%, odd (even+odd,odd+even)  50% => 50/50 3. 3 cubes even (odd (1c+2c)+odd(1c+2c),even(1c+2c)+even(1c+2c))  50%, odd (even(1c+2c)+odd(1c+2c),odd(1c+2c)+even(1c+2c))  50% => 50/50 1/2 (50%/50%) this also correct for any n.
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hmm .. okay i start off considering the different ways that 3 cubes can sum up to an even #
odd, odd, odd
odd, even, even
even, odd, even
even, even, odd
each item gives a probability of 1/8 .. sum them up to get 4/8 = 1/2
is that right ? i wasnt sure whether odd, even, even and even, odd, even should be considered as two separate items



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right
"...with replacement..."  is a key.



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pmenon wrote: hmm .. okay i start off considering the different ways that 3 cubes can sum up to an even #
odd, odd, odd odd, even, even even, odd, even even, even, odd
each item gives a probability of 1/8 .. sum them up to get 4/8 = 1/2
is that right ? i wasnt sure whether odd, even, even and even, odd, even should be considered as two separate items
that's the thing i was after.
thanks



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Re: probabilty [#permalink]
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Updated on: 17 Nov 2007, 17:30
Ravshonbek wrote: 100 cubes were placed into a sack and each numbered from 1 to 100. If three cubes are to be withdrawn with replacement, What is the probability that the sum of the numbers on the withdrawn 3 cubes will be odd.
I get 1/4
two ways to get an odd number...
one odd and two even,
three odd
For each case, the chance is 1/8. Together, 2/8.
Or if we count the permutations  1/2  odd even even, odd odd odd, even odd even, even even odd.
Originally posted by alrussell on 17 Nov 2007, 17:28.
Last edited by alrussell on 17 Nov 2007, 17:30, edited 1 time in total.



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Re: probabilty [#permalink]
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17 Nov 2007, 17:49
alrussell wrote: one odd and two even, three odd
For each case, the chance is 1/8. incorrect. three odd 1/8 one odd and two even 3/8 (odd,even,even), (even,odd,even),(even,even,odd)
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Re: probabilty [#permalink]
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17 Nov 2007, 17:54
walker wrote: alrussell wrote: one odd and two even, three odd
For each case, the chance is 1/8. incorrect. three odd 1/8 one odd and two even 3/8 (odd,even,even), (even,odd,even),(even,even,odd)
As I went on to say, if we are counting all manners in which the numbers can be drawn then the answer is 1/2.



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Why does the order of EVEN, ODD numbers matter when we are looking for sum of the numbers ?
Actually, with replacement makes me think that order does n't matter because the chances of EVEN and ODD numbers are equally likely...
Please explain



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Re: probabilty [#permalink]
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26 Aug 2008, 21:34
Ravshonbek wrote: 100 cubes were placed into a sack and each numbered from 1 to 100. If three cubes are to be withdrawn with replacement, What is the probability that the sum of the numbers on the withdrawn 3 cubes will be odd. OOO combination * probability +OEE combination * probability = 1*(50C1 *50C1*50C1/100C1*100C1*100C1) + 3!/2! *(50C1 *50C1*50C1/100C1*100C1*100C1) = 4 *1/8 =1/2
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Re: probabilty [#permalink]
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26 Aug 2008, 22:22
Odd+Even+Even = Odd
Odd+Odd+Odd = Odd
from 1 to 100, number of even numbered cubes = 50 = number of odd numbered cubes
\((1/50)*(1/50)*(1/49) + (1/50)*(1/49)*(1/48) = (1/50)*(1/49)*[ (1/50) + (1/48) ] =\)



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Re: probabilty [#permalink]
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26 Aug 2008, 22:23
Ravshonbek wrote: 100 cubes were placed into a sack and each numbered from 1 to 100. If three cubes are to be withdrawn with replacement, What is the probability that the sum of the numbers on the withdrawn 3 cubes will be odd. Any integer has the same chance if replaced. so 1/2 should be answer. but with no replacement, it will be different.
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Re: probabilty [#permalink]
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06 Sep 2009, 01:20
from 1 to 100 ,50 no ll be odd and 50 even. if one cube is withdrawn Probability for it to be even P(Even)=50/100=1/2, same way P(odd)=1/2
no lets pick three cubes one by one
first cube can be Odd (O) or Even (E)
if Odd (O) second can again be O or E so two ways would be
OO or OE
in OO case third has to be O to make the sum odd so OOO in OE case third has to be E for the same reason so OEE
now if first cube is Even second can be E or O therefore EE or EO
in EE case third has to be O so EEO in EO case third has to be E so EOE
total sets OOO +OEE+EEO +EOE=1/2*1/2*1/2 + 1/2*1/2*1/2*1/2 + 1/2*1/2*1/2*1/2 + 1/2*1/2*1/2*1/2 =1/8 + 1/8 +1/8 +1/8 =4/8 =1/2



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Re: 100 cubes were placed into a sack and each numbered from 1 [#permalink]
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