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| − | (1-(.001<math>^k</math>)) | + | (1-(.001<math>^k</math>))^365 Serves the condition "atlest" |
| − | Therefore 1-(1-(.001<math>^k</math>)) | + | Therefore 1-(1-(.001<math>^k</math>))^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!! | The above expression should be lesser or eual to 0.001....so solve the equation from thereon!! | ||
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| + | why 1-.001^k I tought that would give probability of being all on. ///-Andrew Hermann | ||
Latest revision as of 05:15, 17 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!!
why 1-.001^k I tought that would give probability of being all on. ///-Andrew Hermann
