18/33?

On Wed, Aug 3, 2011 at 10:42 PM, muthu raj <muthura...@gmail.com> wrote:

> 5)
> Sample space=24C2= 276
> N(S) = 12 C 2 + 12 C 2 =132
>
> Probability= 132/276= 11/23
>
>
> *Muthuraj R
> IV th Year , ISE
> PESIT , Bangalore*
>
>
>
> On Wed, Aug 3, 2011 at 10:36 PM, Prakash D <cegprak...@gmail.com> wrote:
>
>> assume there are 6 1's and 6 0's
>>
>> if two are selected together randomly%
>> the possible outcomes are 00, 11, 01, 10
>>
>> 00- 30 possibilities
>> 11 - 30 possibilities
>>
>> 10 - 36 possibilities
>> 01 - 36 possibilities
>>
>>
>> probability of 10  or 01 =  (36+36)/(30+30+36+36)
>>
>> =18/31
>>
>>  is it one of the options?
>>
>>
>> On Wed, Aug 3, 2011 at 10:28 PM, JAIDEV YADAV <jaid...@gmail.com> wrote:
>>
>>> there is no error in first question ... u thik so.. correct me if wrong
>>> ...
>>>
>>>
>>> On Wed, Aug 3, 2011 at 10:26 PM, Kamakshii Aggarwal <
>>> kamakshi...@gmail.com> wrote:
>>>
>>>> i think dere is nothing wrong in ques 1.is there any error?
>>>>
>>>>
>>>> On Wed, Aug 3, 2011 at 9:46 PM, cegprakash <cegprak...@gmail.com>wrote:
>>>>
>>>>> 2 is pretty easy..
>>>>>
>>>>> 3: both
>>>>>
>>>>> 5: 6/12 * 6/11  = 3/11
>>>>>
>>>>> am i right?
>>>>>
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>>>>
>>>>
>>>> --
>>>> Regards,
>>>> Kamakshi
>>>> kamakshi...@gmail.com
>>>>
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>>> JaiDev Yadav
>>> (National Yoga Champion)
>>> Computer Engg. Dept.
>>> National Institute of Technology
>>> Kurukshetra,Haryana
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