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if x has a poisson distribution with u=3, what is the probability that x=3
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\[P(x=3)=\frac{3^3e^{-3}}{3!}\]if i remember correctly
ok that was wrong general form is \[P(x=k)=\frac{\lambda^ke^{-\lambda}}{k!}\]
compute via a calculator, although \[\frac{3^3}{3!}=\frac{3^2}{2}=\frac{9}{2}\] you will still need a calculator for \(e^{-3}\) though
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