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]> 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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