@shady: i'll explain it clearly
the possible outcomes after a blue is chosen are
blue black
packet1 1 0
packet2 1 1
packet3 0 2
and
blue black
packet1 2 0
packet2 0 1
packet3 0 2
prob of choosing a blue from 1st possibility = 1/3*1 + 1/3*1/2 + 1/3*0 =
1/2
prob of choosing a blue from 2nd possibility= 1/3* 1 + 1/3*0 + 1/3* 0 = 1/3
so totally there are 5/6 possibilities
On Mon, Aug 8, 2011 at 12:05 AM, shady <[email protected]> wrote:
> no it is 1/2
>
>
> On Sun, Aug 7, 2011 at 9:14 PM, Prakash D <[email protected]> wrote:
>
>> The solution is 5/6 for sure
>>
>>
>> the possible outcomes after a blue is chosen are
>>
>> blue 1 1 0
>> black 0 1 2
>>
>> and
>>
>> blue 2 0 0
>> black 0 1 2
>>
>> prob of choosing a blue from 1st possibility = 1/3+1/6
>>
>> prob of choosing a blue from 2nd possibility= 1/3
>>
>> so totally there are 5/6 possibilities
>>
>>
>>
>> On Sun, Aug 7, 2011 at 3:05 PM, Kunal Yadav <[email protected]>wrote:
>>
>>> Agree with puneet completely. After first blue pen, only possible
>>> outcomes are blue or black and hence 1/2.
>>>
>>>
>>> On Sat, Aug 6, 2011 at 12:28 AM, Puneet Goyal
>>> <[email protected]>wrote:
>>>
>>>>
>>>> @shiv : you are considering the case when you need to know the
>>>> probability of both pens being blue, but in the question you already know
>>>> that 1st one is blue so you dont need to care about it, and also the 3rd
>>>> packet getting eliminated coz of it increases the probability to 1/2
>>>>
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>>>
>>>
>>>
>>> --
>>> Regards
>>> Kunal Yadav
>>> (http://algoritmus.in/)
>>>
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>>
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