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
On Mon, Sep 19, 2011 at 1:04 PM, Nitin Garg <[email protected]>wrote:
> Question 3 -
> To eliminate one player, you need to host atleast 2 matches and make him
> loose in both 2. These 2 matches can not contribute to elimination of any
> other player.
> So, min 2 matches for every player who is to be eliminated, hence 100.
>
>
> On Mon, Sep 19, 2011 at 11:54 AM, Bhanu Chowdary
> <[email protected]>wrote:
>
>> @Nitin: Answer to question 3 is 50.
>>
>>
>> On Mon, Sep 19, 2011 at 11:44 AM, praveen raj <[email protected]>wrote:
>>
>>> @nitin Plz explain how u have reached answer of question no. 4 and 6
>>>
>>> On 19-Sep-2011 12:26 AM, "Nitin Garg" <[email protected]> wrote:
>>> >
>>> > Answer 3 - 100
>>> > Answer 6 - 103
>>> > Answer 4 - 194 total processes including the parent
>>> > Answer 7 - 12 km south, 12 km east
>>> >
>>> >
>>> > On Sun, Sep 18, 2011 at 11:53 PM, Ashima . <[email protected]>
>>> wrote:
>>> >>
>>> >> @malay: how cm n+logn-2?
>>> >> cn u explain the logic ?
>>> >>
>>> >> Ashima
>>> >> M.Sc.(Tech)Information Systems
>>> >> 4th year
>>> >> BITS Pilani
>>> >> Rajasthan
>>> >>
>>> >>
>>> >>
>>> >>
>>> >> On Sun, Sep 18, 2011 at 11:07 AM, Ashima . <[email protected]>
>>> wrote:
>>> >>>
>>> >>> rite! 62.5%
>>> >>>
>>> >>> Ashima
>>> >>> M.Sc.(Tech)Information Systems
>>> >>> 4th year
>>> >>> BITS Pilani
>>> >>> Rajasthan
>>> >>>
>>> >>>
>>> >>>
>>> >>>
>>> >>> On Sat, Sep 17, 2011 at 9:04 PM, malay chakrabarti <
>>> [email protected]> wrote:
>>> >>>>
>>> >>>> create a tournament tree.in each round one value is eliminated to
>>> obtain in the process the winner or the highest value in n-1 comparisons.
>>> Then check the queue of the winner which contains log(n) entries of the
>>> values beaten by the winner which implicitly will contain the runners
>>> up.Then log(n)-1 comparisons to find the highest among all the losers whom
>>> the winner had beaten. So all over complexity will be n-1 +log(n) -1 =
>>> n+log(n)-2. Hp that answers ur query. nice question btw :)
>>> >>>>
>>> >>>>
>>> >>>> On Sun, Sep 18, 2011 at 8:02 AM, VIHARRI <[email protected]>
>>> wrote:
>>> >>>>>
>>> >>>>> hey i'm also thinking n + logn -2.. but couldnt able to figure out
>>> >>>>> how??? can you please explain the logic
>>> >>>>>
>>> >>>>> --
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>>> >>>>>
>>> >>>>
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>>> >>>
>>> >>>
>>> >>
>>> >> --
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>>> >
>>> >
>>> >
>>> >
>>> > --
>>> > Nitin Garg
>>> >
>>> > "Personality can open doors... but only Character can keep them open"
>>> >
>>> > --
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>>
>>
>>
>> --
>> Bhanu Chowdary
>>
>> --
>> You received this message because you are subscribed to the Google Groups
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>>
>
>
>
> --
> Nitin Garg
>
> "Personality can open doors... but only Character can keep them open"
>
--
Nitin Garg
"Personality can open doors... but only Character can keep them open"
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