total possible outcomes= 6*6*6*6
possibility of sum gives 2 in both pair --> 1*1*1*1 =1
possibility of sum gives 3 in both pair --> 2*1*2*1 =4
{because the possibilities are (1,2)(1,2), (1,2)(2,1), (2,1)(1,2),
(2,1)(2,1)}
possibility of 4 --> 9 { 2,2 1,3 and 3,1}
possibility of 5 --> 16 {1,4 2,3 3,2 4,1}
possibility of 6 --> 25 {1,5 2,4 3,3 4,2 5,1}
possibility of 7 --> 36 {1,6 2,5 3,4 4,3 5,2 6,1}
possibility of 8 --> 25 {2,6 3,5 4,4 5,3 6,2}
9--> 16
10 --> 9
11 --> 4
12 --> 1
so probability required = (1+4+9+16+25+36+25+16+9+4+1)/(6*6*6*6)
=146/1296
=73/648
On Fri, Aug 12, 2011 at 1:19 AM, Hurricane <[email protected]> wrote:
> Let f(x)= no. of ways of getting a sum x when pair of dice is thrown.
>
> S()=sum .
>
> tot=S( f(i) ) 2<=i<=12
>
> So the solution is : S( f(i)*(tot-f(i)) )/(6^4).
>
>
> Is daT fine.. ?
>
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