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- Apr 7th 2009, 04:13 PMsteffan_09Poisson
- Apr 7th 2009, 05:41 PMmatheagle
(a) $\displaystyle {e^{-4}4^6\over 6!} $.

(b) $\displaystyle P(X\ge 10) =1-P(X\le 9)$ where $\displaystyle \lambda =8$ since this is a two week period and you expect twice as many parts used.

So, $\displaystyle P(X\ge 10) =1-\sum_{k=0}^9 {e^{-8}8^k\over k!} $.

(c) Here you want three 'successes' in three weeks (binomial) where p is from (a)...$\displaystyle \biggl({e^{-4}4^6\over 6!}\biggr)^3 $.

Sorry, I do not understand (d).