@Nitin Garg Question 6 -
i agree that greater the sum is and greater the probability to getting it. but in given question if sum>100 then rolling is stopped so for P(106)=P(100)*1/6 P(105)=P(100)*1/6+P(99)*1/6 . . . P(101)=P(100)*1/6+P(99)*(1/6)+P(98)*(1/6)+P(97)*(1/6)+..+P(95)*(1/6) now P(101) is more cleare me if something is wrong. On Mon, Sep 19, 2011 at 1:35 PM, Nitin Garg <[email protected]>wrote: > Question 6 - > Intuitively you can see that the greater the sum is, the greater the > favorable events in sample space. > > e.g. - sum = 1 .. cases {(1)} Pr = 1/6 > sum = 2 cases {(2),(1,1)} Pr = 1/6 + 1/36 > sum = 3 cases {(3),(2,1)(1,2)(1,1,1)} Pr = 1/6 + 1/36 +1/36 + > 1/216 > > > for a more formal proof, look at the recursion - > > > P(k) = (P(k-6) + P(k-5) + P(k-4)... P(k-1)))/6 > > where P(0) = 1, P(i) = 0 for i<0 > > Base case - > P(2) > P(1) > > Hypothesis - > > P(i) > P(i-1) for all i <= k > > To prove > P(k+1) > P(k) > > Proof > P(k+1) - P(k) = (P(k) - P(k-6))/6 > 0 > > > > -- > Pankaj Agarwal > Communication and Computer Engineering > LNMIIT,jaipur > > -- 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?hl=en.
