A secretary makes 2 errors per page on average. What is the probability that he makes
(a)4 or more errors on the next page he types?
(b)no errors on the next page he types?
This is a Poisson distribution problem, with $\displaystyle \lambda = 2$
a) $\displaystyle P(X \geq 4) = 1 - P(X < 4) $
You know the basic formula for the Poisson distribution:
$\displaystyle \frac{\lambda ^{k} e ^{- \lambda}}{k!}$
$\displaystyle = 1 - \frac{2 ^{0} e ^{- 2}}{0!} -$ $\displaystyle \frac{2 ^{1} e ^{- 2}}{1!} -$ $\displaystyle \frac{2 ^{2} e ^{- 2}}{2!} -$ $\displaystyle \frac{2 ^{3} e ^{- 2}}{3!} $
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b) $\displaystyle \frac{2 ^{0} e ^{- 2}}{0!}$