Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 30 Jul 2016, 14:20

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
VP
Joined: 09 Jul 2007
Posts: 1104
Location: London
Followers: 6

Kudos [?]: 97 [0], given: 0

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

### Show Tags

17 Nov 2007, 16:40
1
This post was
BOOKMARKED
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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 If three cubes are to be withdrawn with replacement, What is the probability that the sum of the numbers on the withdrawn 3 cubes will be odd.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
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 490

Kudos [?]: 2931 [0], given: 360

### Show Tags

17 Nov 2007, 17:05
Expert's post
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.
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

SVP
Joined: 28 Dec 2005
Posts: 1575
Followers: 3

Kudos [?]: 128 [0], given: 2

### Show Tags

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
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 490

Kudos [?]: 2931 [0], given: 360

### Show Tags

17 Nov 2007, 17:20
Expert's post
right

"...with replacement..." - is a key.
VP
Joined: 09 Jul 2007
Posts: 1104
Location: London
Followers: 6

Kudos [?]: 97 [0], given: 0

### Show Tags

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
Followers: 1

Kudos [?]: 2 [0], given: 0

### Show Tags

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.
CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 490

Kudos [?]: 2931 [0], given: 360

### Show Tags

17 Nov 2007, 17:49
Expert's post
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)
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Manager
Joined: 08 Nov 2007
Posts: 99
Followers: 1

Kudos [?]: 2 [0], given: 0

### Show Tags

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
Followers: 0

Kudos [?]: 14 [0], given: 0

### Show Tags

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...

SVP
Joined: 07 Nov 2007
Posts: 1820
Location: New York
Followers: 30

Kudos [?]: 748 [0], given: 5

### Show Tags

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
_________________

Smiling wins more friends than frowning

Director
Joined: 03 Sep 2006
Posts: 879
Followers: 6

Kudos [?]: 605 [0], given: 33

### Show Tags

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

$$(1/50)*(1/50)*(1/49) + (1/50)*(1/49)*(1/48) = (1/50)*(1/49)*[ (1/50) + (1/48) ] =$$
SVP
Joined: 29 Aug 2007
Posts: 2492
Followers: 65

Kudos [?]: 682 [0], given: 19

### Show Tags

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.
_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

Manager
Joined: 05 Jun 2009
Posts: 111
Followers: 3

Kudos [?]: 145 [0], given: 4

### Show Tags

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
Display posts from previous: Sort by