Past records show that 80% of the operations performed by a certain doctor were successful . If he performs 4 operations in a day what is the probability that at least 3 operations will be successful ?

Answer

`(512)/(625)`

Explanation

`p=(80)/(100) =(4)/(5) ,q=(1-(4)/(5)) =(1)/(5) " and " n=4`
`P(X=r) =.^(n)C_(r ).p^(r ).q^((n-r)) =.^(4)C_(r ).((4)/(5))^(r ) .((1)/(5))^((4-r))`
`P(Xge 3) =P(X=3) +P(X=4)`
`=[.^(4)C_(3 ).((4)/(5))^(3).((1)/(5))+.^(4)C_(4).((4)/(5))^(4)]`