Hi! Assume 5000 test cases of A,B,C,D.. and The probabilities are as mentioned above.
Then If loop will fail for 75% of test cases and else will be executed.. In that 75% test cases, only 75% test cases will satisfy the inner if statement.. So 3/4 * 3/4 = 9/16. 9/16 * 5000 = 2812.. I guess that 'll do. :) On Dec 14, 10:05 pm, Saurabh Koar <saurabhkoar...@gmail.com> wrote: > @Sravan: Plz explain the logic.. > > On 12/15/10, Sravan Akepati <sravan.akep...@gmail.com> wrote: > > > > > > > > > (1-0.25)* 0.75*5000 = 2812.5 > > > On Wed, Dec 15, 2010 at 9:31 AM, ankit sablok <ankit4...@gmail.com> wrote: > > >> well i still believe that the calling of foo2 is independent plzzz suggest > >> me the solution if i am wrong a detailed one thanx in advance > > >> On Wed, Dec 15, 2010 at 1:22 AM, Saurabh Koar > >> <saurabhkoar...@gmail.com>wrote: > > >>> The function foo2 will be called iff the condition if(C<D) evaluates to > >>> be > >>> true. > >>> Given that C<D turns out to be true 75% times.So why the call to foo2 > >>> will be independent?? > >>> I think it is only the simple math.Correct me if I am wrong.. > > >>> On 12/15/10, ankit sablok <ankit4...@gmail.com> wrote: > >>> > what i think is that the number of times foo2 being called is > >>> > independent of the percentages given in the question it may be called > >>> > 5000 times or 4999 times and continuinf in this fashion also none of > >>> > the times as in every case there's 1/4 probability of A<B and 3/4 of > >>> > C<D so as per me we cannot decide givn the percentage of success and > >>> > failure any suggestions are always welcomed > > >>> > On Dec 15, 12:06 am, bittu <shashank7andr...@gmail.com> wrote: > >>> >> void foo1() > >>> >> { > >>> >> if(A<B) > >>> >> Then {_/*.... */} > >>> >> else > >>> >> if(C<D) > >>> >> then foo2() > > >>> >> } > > >>> >> How many time foo2() would get called given > >>> >> A<B 25% of the times and C<D 75% of the times and foo1() is called > >>> >> 5000 times > > >>> >> although i had diff...solution..but i wants to confirm wid others..so > >>> >> hav a look > > >>> >> Regards > >>> >> Shashank Mani > >>> >> BIT Mesra > > >>> > -- > >>> > You received this message because you are subscribed to the Google > >>> Groups > >>> > "Algorithm Geeks" group. > >>> > To post to this group, send email to algoge...@googlegroups.com. > >>> > To unsubscribe from this group, send email to > >>> > algogeeks+unsubscr...@googlegroups.com<algogeeks%2bunsubscr...@googlegroups > >>> > .com> > >>> . > >>> > For more options, visit this group at > >>> >http://groups.google.com/group/algogeeks?hl=en. > > >>> -- > >>> You received this message because you are subscribed to the Google Groups > >>> "Algorithm Geeks" group. > >>> To post to this group, send email to algoge...@googlegroups.com. > >>> To unsubscribe from this group, send email to > >>> algogeeks+unsubscr...@googlegroups.com<algogeeks%2bunsubscr...@googlegroups > >>> .com> > >>> . > >>> For more options, visit this group at > >>>http://groups.google.com/group/algogeeks?hl=en. > > >> -- > >> You received this message because you are subscribed to the Google Groups > >> "Algorithm Geeks" group. > >> To post to this group, send email to algoge...@googlegroups.com. > >> To unsubscribe from this group, send email to > >> algogeeks+unsubscr...@googlegroups.com<algogeeks%2bunsubscr...@googlegroups > >> .com> > >> . > >> For more options, visit this group at > >>http://groups.google.com/group/algogeeks?hl=en. > > > -- > > You received this message because you are subscribed to the Google Groups > > "Algorithm Geeks" group. > > To post to this group, send email to algoge...@googlegroups.com. > > To unsubscribe from this group, send email to > > algogeeks+unsubscr...@googlegroups.com. > > For more options, visit this group at > >http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.