Sorry, I mis understood the ques.

On Sat, Aug 6, 2011 at 8:56 PM, Pankaj <[email protected]> wrote:

> There are two cases: First if we get even number in first roll.
> So P(A)= 1/2
> And if We get odd number in second roll, then we need keep rolling untill
> we find another odd number.
> So, (1/2)*[1/2] + 1/2*1/2*1/2 .... upto infinity.
>
>
> On Sat, Aug 6, 2011 at 8:46 PM, Nitin Gupta <
> [email protected]> wrote:
>
>> 1/6*5/6*5/6*3+ 1/6*1/6*5/6*3+ 1/6*1/6*1/6 =91/216
>>
>>
>> On Sat, Aug 6, 2011 at 8:24 PM, muthu raj <[email protected]> wrote:
>>
>>> Microsoft written:
>>>
>>> What is the probability of getting atleast one 6 in  3 attempts of a
>>> dice?
>>>
>>>
>>> *Muthuraj R
>>> IV th Year , ISE
>>> PESIT , Bangalore*
>>>
>>>
>>>
>>>
>>> On Sat, Aug 6, 2011 at 7:34 AM, shady <[email protected]> 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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>
>

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