How do you get P(0.25 being correct) = P(0.5 being correct) = P(0.6 being 
correct) = 1/3?

On Saturday, September 8, 2012 4:42:51 PM UTC+5:30, Shruti wrote:
>
> shouldn't it b done like this : 
>
> P(correct answer when choosing randomly )= P(choosing 0.25)*P(0.25 being 
> correct ans)+ P(choosing 0.60)*P(0.60 being correct ans) + P(choosing 
> 0.50)*P(0.50 being correct ans) 
>
>
> = 2/4*1/3  +  1/4*1/3  +  1/4*1/3     [since 3 possible answers, hence 
> prob of an ans being correct=1/3]
>
> taking 1/3 common,
> =1/3(2/4+1/4+1/4)
> =1/3
>
> pls correct me if i'm wrong
>
> On Fri, Sep 7, 2012 at 7:54 PM, isandeep <[email protected] <javascript:>
> > wrote:
>
>> Ans : 0.5
>>
>> there is two cases :
>>
>> i) if correct ans is 0.25
>>     probability will be 2/4 = 0.5
>> ii) if correct ans is 0.6 or 0.50
>>     probability will be 1/3 = 0.33 
>>
>> since 0.33 is not in option so correct answer will be 0.5.
>>
>> On Friday, September 7, 2012 6:05:08 PM UTC+5:30, noname wrote:
>>>
>>> [image: -]14Answers <http://www.careercup.com/question?id=14553727>
>>>
>>> What is the probability of being the answer correct for this question, 
>>> when the answer is chosen randomly:
>>>
>>> a. 0.25
>>> b. 0.60
>>> c. 0.25
>>> d. 0.50
>>>  
>>>
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