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# 100 cubes were placed into a sack and each numbered from 1

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VP
Joined: 09 Jul 2007
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Location: London
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 If three cubes are to be withdrawn with replacement, What is the probability that the sum of the numbers on the withdrawn 3 cubes will be odd.
CEO
Joined: 17 Nov 2007
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Concentration: Entrepreneurship, Other
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17 Nov 2007, 17:05
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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SVP
Joined: 28 Dec 2005
Posts: 1558

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17 Nov 2007, 17:16
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
CEO
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Concentration: Entrepreneurship, Other
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17 Nov 2007, 17:20
right

"...with replacement..." - is a key.
VP
Joined: 09 Jul 2007
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Location: London

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17 Nov 2007, 17:24
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
Manager
Joined: 08 Nov 2007
Posts: 99

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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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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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Manager
Joined: 08 Nov 2007
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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.
Intern
Joined: 11 Jun 2007
Posts: 22

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19 Nov 2007, 19:58
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...

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Joined: 07 Nov 2007
Posts: 1799
Location: New York

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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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Director
Joined: 03 Sep 2006
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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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Joined: 29 Aug 2007
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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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Manager
Joined: 05 Jun 2009
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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
Re: probabilty   [#permalink] 06 Sep 2009, 01:20
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