@ Minjie Zha ,
hey, how does these things clicks to u , as i thought it for 2 hrs.
and still couldn't find a completely correct sol..

On Sep 19, 2:04 pm, Minjie Zha <[email protected]> wrote:
> Oh yes, I made a mistake.
> Your are right.
>
> On Sep 18, 12:02 am, ashish gupta <[email protected]> wrote:
>
>
>
> > i think there might be some modification
>
> > On Thu, Sep 17, 2009 at 4:17 PM, Minjie Zha <[email protected]> wrote:
>
> > > Let PH(j,w) be the probability of getting w heads from 1...j coins,
> > > 0<=j<=k, 0<=w<=k.
> > > So we have:
> > > PH(0,0) = 1
>
> >  PH( j, w ) = 0  if w< 0
>
> > > PH(0,w) = 0 for w>0
> > > PH(j,0) = (1-P(1))(1-P(2))...(1-P(j))
>
> > > PH(j,w) = PH(j-1,w) + PH(j-1,w-1)PH(j)
>
> > and equation should be
> > PH(j, w)  = PH(j-1,w) (1-P(j)) + PH( j-1, w-1) PH(j)
>
> > pls correct if i am wrong...
>
> > --
> > ashish
>
> > > Any comments?
>
> > > On Sep 9, 5:50 pm, Nagendra Kumar <[email protected]> wrote:
> > > > @all:
> > > >           There are k baised coins with probabilty of coming head is
> > > > P(i)  i = 1 to k.  If all these coins are  tossed together. find the
> > > > probabilty of getting i heads ( i < = k).
> > > >    think in Dynamic Programming.
> > > > -Nagendra

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