(New page: Take P,The probability of not getting connected on one day = (0.001)<math>^k</math>)
 
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Take P,The probability of not getting connected on one day = (0.001)<math>^k</math>
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  Take P,The probability of not getting connected on any one day = (0.001)<math>^k</math>
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  (1-(.001<math>^k</math>))<math>^365</math> Serves the condition "atlest"
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Therefore  1-(1-(.001<math>^k</math>))<math>^365</math> Tells us the proability of total outrage happening "atleast" once a year.
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The above expression should be lesser or eual to 0.001....so solve the equation from thereon!!

Revision as of 19:13, 16 September 2008

 Take P,The probability of not getting connected on any one day = (0.001)$ ^k $


  (1-(.001$ ^k $))$ ^365 $ Serves the condition "atlest" 
Therefore  1-(1-(.001$ ^k $))$ ^365 $ Tells us the proability of total outrage happening "atleast" once a year. 

The above expression should be lesser or eual to 0.001....so solve the equation from thereon!!

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