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(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) 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 --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/algogeeks -~----------~----~----~----~------~----~------~--~---
