Hey sry for my above post. I got a little confused. x is the no of times
dice is rolled so
e(x)=e(1)+e(2)+e(3)+....
=1/2 + 2*1/(2*2) + 3*1/(2*2*2) + ......
Please correct me if m wrong..
On Sat, Aug 6, 2011 at 10:54 PM, Kunal Yadav <kunalyada...@gmail.com> wrote:
> Expected value of a random variable x is defined as E(x)= summation of
> xp(x) over all value of x where p(x) is the probability.
> so in this case
> E(x)= E(2)+E(4)+ E(6)+ .....
> = 2*1/6 + 4* 3/(6*6)+ 6*10/(6*6*6) + .....
>
>
> On Sat, Aug 6, 2011 at 9:19 PM, sukran dhawan <sukrandha...@gmail.com>wrote:
>
>>
>>
>> On Sat, Aug 6, 2011 at 8:24 PM, muthu raj <muthura...@gmail.com> wrote:
>>
>>> Microsoft written:
>>>
>>> What is the probability of getting atleast one 6 in 3 attempts of a
>>> dice?
>>>
>>> probability of not getting 6 = 5/6 * 5/6 * 5/6 = 91/216
>>>
>> so ans 1- 91/216
>> let me know how was the paper and the pattern
>>
>>> *Muthuraj R
>>> IV th Year , ISE
>>> PESIT , Bangalore*
>>>
>>>
>>>
>>>
>>> On Sat, Aug 6, 2011 at 7:34 AM, shady <sinv...@gmail.com> wrote:
>>>
>>>> Hi,
>>>>
>>>> A fair dice is rolled. Each time the value is noted and running sum is
>>>> maintained. What is the expected number of runs needed so that the sum is
>>>> even ?
>>>> Can anyone tell how to solve this problem ? as well as other related
>>>> problems of such sort....
>>>>
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>>>
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>>
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>
>
>
> --
> Regards
> Kunal Yadav
> (http://algoritmus.in/)
>
>
--
Regards
Kunal Yadav
(http://algoritmus.in/)
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