On 23/09/2009, vicky <mehta...@gmail.com> wrote: > > @ 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 <diego...@gmail.com> wrote: >> Oh yes, I made a mistake. >> Your are right. >> >> On Sep 18, 12:02 am, ashish gupta <ashish12.it...@gmail.com> wrote: >> >> >> >> > i think there might be some modification >> >> > On Thu, Sep 17, 2009 at 4:17 PM, Minjie Zha <diego...@gmail.com> 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 <nagendra....@gmail.com> 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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