I was trying to solve this.. was somewhat close before I looked at ur
solution.
But.. equal probability for 1-12 (1 inclusive is a good hint for putting on
0's)
On Saturday, 7 July 2012 15:58:35 UTC-4, Abhishek Sharma wrote:
@Raj: trace karke dekh na yaar when u have 3 0s and 3 6s.. the
You are given 2 dice. Both are fair. One of the dice has no numbers printed
on it. You have to label the unmarked dice such that when both the dice are
thrown, the sum on the faces is evenly distributed between 1 and 12 .
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plz explain ur ans .how it is calculated?
On Sat, Jul 7, 2012 at 12:54 PM, Amol Sharma amolsharm...@gmail.com wrote:
Three 0's and Three 6's.
36 possibilities and 12 possible values of sum, so each sum should come in
3 possibilities and hence three 0's and three 6's.
--
Amol Sharma
Label 3 of the faces with 0 and other 3 faces with 6.
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Prakhar Jain
IIIT Allahabad
B.Tech IT 3rd Year
Mob no: +91 9454992196
E-mail: rit2009...@iiita.ac.in
jprakha...@gmail.com
On Sat, Jul 7, 2012 at 12:52 AM, Hraday Sharma hradaysha...@gmail.comwrote:
You are given 2 dice.
@Raj: trace karke dekh na yaar when u have 3 0s and 3 6s.. the sum
distribution would look like this:
given below are the possibilities:
Combination of 1,2,3,4,5,6 with 0
1+0 = 1
2+0 = 2
3+0 = 3
OR
4+0 = 4
5+0 = 5
6+0 = 6
Combination of 1,2,3,4,5,6 with 6
1+6 = 7
2+6 = 8
3+6 = 9
OR
4+6
