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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