I am doing this question for homework and I've arrived at a solution which does not match the book and I am wondering what I'm doing wrong. I feel like my approach is not that much different but we get relatively different answers. Let me know if you find out my flaw, or perhaps there is none.

A certain typing agency employs 2 typists. The average number of errors per article is 3 when typed by the first typist and 4.2 by the second. If your article is equally likely to be typed by either typist, approximate the probability that it will have no errors.

Here is my approach:

Let $X =$ # errors, where $X$ is roughly Poisson. Now let

$Y = \begin{cases} 0 & \text{if first typist} \\ 1 & \text{if second typist} \end{cases}$

Then we have:

$EX = E[E(X|Y)] = E(X|Y=0)P(Y=0) + E(X|Y=1)P(Y=1) = \frac{1}{2}(3 + 4.2) = 3.6$.

Then since X is Poisson we know that: $P(X = 0) = \frac{3.6^0e^{-3.6}}{0!} = e^{-3.6} = \fbox{0.027.}$ However, the book gives $0.032$. Not sure what I am doing wrong.

  • 1
    $\begingroup$ $$P(X=0)=\sum_yP(X=0\mid Y=y)P(Y=y)=e^{-\lambda_0}\tfrac12+e^{-\lambda_1}\tfrac12$$ $\endgroup$ – Did Apr 15 '17 at 22:07

$X$ is not poisson. It is a mixture of two poisson distributions. So your step when you use the (correct) mean of $3.6$ and compute $P(X=0)$ assuming $X$ is a Poisson with mean $3.6$ is wrong.

Instead, use total probability, same way you computed the mean, but with the entire distribution. So you have $$ P(X=0)= P(X = 0|Y=0)P(Y=0) + P(X=0|Y=1)P(Y=1) = e^{-3}\frac{1}{2} + e^{-4.2}\frac{1}{2}=.032 $$

  • $\begingroup$ Yes this makes sense, thank you! $\endgroup$ – student_t Apr 15 '17 at 22:24

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