Past experience shows that 80 % of the operations performed by a doctor arc successful. If he performs 4 operations in a day, what is the probability that at least three operations will be successful?

Solution

Here n = 4
$straight p space equals space 80 over 100 space equals space 4 over 5 comma space space space space straight q space space equals 1 space minus space straight p space equals space 1 minus 4 over 5 space equals space 1 fifth$
P(at least three successful operations) = P(3) + P(4)
$equals space straight C presuperscript 4 subscript 3 space open parentheses 1 fifth close parentheses space open parentheses 4 over 3 close parentheses cubed space plus space straight C presuperscript 4 subscript 4 space open parentheses 4 over 5 close parentheses to the power of 4 space equals 4 over 1 cross times 1 fifth cross times 64 over 125 plus 1 cross times 256 over 625 equals space 256 over 625 plus 256 over 625 plus 512 over 625 equals space 0.8192.$

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Probability

Past experience shows that 80 % of the operations performed by a doctor arc successful. If he performs 4 operations in a day, what is the probability that at least three operations will be successful?

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