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# Find $P(X<3)$ and E(X).

I am a little confused with finding $$P(X<3)$$ and $$E(X)$$. Here, A,B,C are Poisson random variables with parameters 2.6,3, and 3.4, respectively. Let X be the number of errors typed in a manuscript. Each of three typists A,B,C are equally likely to type the manuscript with error rates given. For the

$$E(X)=\frac{1}{3}(2.6+3.4+3)=3$$

but I am not sure. I am, also, having problems understanding $$P(X<3)$$ I am getting

\begin{align*}P(X<3) &= \left(\frac{1}{3}\right)\left(e^{-3.4}\frac{3.4^3}{3!}+e^{-2.6}\frac{2.6^3}{3!}+e^{-3}\frac{3^3}{3!}\right)\\&+\left(\frac{1}{3}\right)\left(e^{-3.4}\frac{3.4^2}{2!}+e^{-2.6}\frac{2.6^2}{2!}+e^{-3}\frac{3^2}{2!}\right)\\&+\left(\frac{1}{3}\right)\left(e^{-3.4}\frac{3.4^1}{1!}+e^{-2.6}\frac{2.6^1}{1!}+e^{-3}\frac{3^1}{1!}\right)\\&+\left(\frac{1}{3}\right)\left(e^{-3.4}\frac{3.4^0}{0!}+e^{-2.6}\frac{2.6^0}{0!}+e^{-3}\frac{3^0}{0!}\right)\\&=0.64719.\end{align*}

but I feel is wrong. Can someone explain this to me?