(1-0.25)* 0.75*5000 = 2812.5

On Wed, Dec 15, 2010 at 9:31 AM, ankit sablok <[email protected]> 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 <[email protected]>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 <[email protected]> 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 <[email protected]> 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
>> >
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