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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 If three cubes are to be withdrawn with replacement, What is the probability that the sum of the numbers on the withdrawn 3 cubes will be odd.



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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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17 Nov 2007, 17:28
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 If three cubes are to be withdrawn with replacement, What is the probability that the sum of the numbers on the withdrawn 3 cubes will be odd.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.
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










