it' probability , how do we define best or worst in probability ?

On Wed, Dec 15, 2010 at 12:25 PM, Kathir <[email protected]> wrote:
> I think, we must give answers in the range..
>
> See, what happens if the
>
> Test cases which doesn't satisfy (A<B) can be the test cases which
> satisfy the condition (C<D)
>
> In that scenario,
>
> !(A<B)  and  (C<D) co-occurs,  Ouput will be 3/4 *1  * 5000 = 3750..
>
> o.w,  Worst case , 3/4 * 2/3 * 5000 = 2500 ..
>
> Range - 2500 < x < 3750
>
>
>
> Correct me if i am wrong.. :)
>
>
> On Dec 14, 10:37 pm, Ankur Khurana <[email protected]> wrote:
>> i think sravan is right. we go to C<D condition after A<B have failed.
>> For that to happen, we have prob. of .75 . now the probability of C<D
>> is .75 . so total probability is .75 * .75 as both are mutualy
>> exclusive events. so total no. foo2 is called is 2812.5 .
>>
>> On Wed, Dec 15, 2010 at 11:55 AM, Ankur Murarka
>>
>>
>>
>>
>>
>>
>>
>> <[email protected]> wrote:
>> > @Sravan
>> > There seems to be a little problem in your solution. Your are probably
>> > assuming that 75% of C is less than D after the condition that A is greater
>> > than B while thats not the case according to the question.
>>
>> > My Solution -
>> > Out of 5000 cases, A>B in 3750 of them and C<D in 3750 of them again. Thus,
>> > foo2 should run atleast 2500 times and not more than 3750 times depending 
>> > on
>> > the input combinations.
>>
>> > On Wed, Dec 15, 2010 at 11:35 AM, Saurabh Koar <[email protected]>
>> > wrote:
>>
>> >> @Sravan: Plz explain the logic..
>>
>> >> On 12/15/10, Sravan Akepati <[email protected]> wrote:
>> >> > (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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