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

Question

I am a little confused with finding $P(X<3)$ and $E(X)$. I ask this question before but no one answer. 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:

$\begin{align*}P(X<3) &= \left(\frac{1}{3}\right)\left(e^{-3.4}\frac{3.4^2}{2!}+e^{-2.6}\frac{2.6^2}{2!}+e^{-3}\frac{3^2}{2!}\right)\\&+\left(\frac{1}{3}\right)\left(e^{-3.4}\frac{3.4^1}{1!}+e^{-2.6}\frac{2.6^1}{1!}+e^{-3}\frac{3^1}{1!}\right)\\&+\left(\frac{1}{3}\right)\left(e^{-3.4}\frac{3.4^0}{0!}+e^{-2.6}\frac{2.6^0}{0!}+e^{-3}\frac{3^0}{0!}\right)\\&=0.42711.\end{align*}$

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

Answer 1

You have to use the law of Total Probability.

Recall that $E[X] = \sum_{k=1}^\infty kP(X = k)$. We write : $$ P(X = k) = P(X = k \mid A)P(A) + P(X = k \mid B)P(B) + P(X = k \mid C) P(C) $$

where $P(X = k \mid A)$ is the probability that $A$ makes exactly $k$ errors (similarly for $B,C$) and $P(A)$ is the probability that $A$ is chosen to type (similarly $B,C$).

From here, we also get : \begin{align} E[X] &= \sum_{k=1}^\infty k[ P(X = k \mid A)P(A) + P(X = k \mid B)P(B) + P(X = k \mid C)P(C)] \\ &= \frac 13\sum_{k=1}^\infty k[P(X = k \mid A) + P(X = k\mid B) +P(X = k\mid C)] \\ &= \frac 13\left(\sum_{k=1}^\infty kP(X = k\mid A) + \sum_{k=1}^\infty kP(X = k\mid B)+\sum_{k=1}^\infty kP(X = k\mid C)\right) \\ &= \frac 13\left(E[X \mid A] + E[X\mid B] + E[X \mid C]\right) \\ &= \frac 13(2.6+3.4+3) = 3 \end{align}

What you've done for the second part is also correct and justified by : $$ P(X < 3) = P(X =0) + P(X = 1) + P(X = 2) \\ = \sum_{i=0}^2 P(X = i | A)P(A) + P(X = i|B)P(B) + P(X = i|C)P(C) $$

Which is exactly what you've computed above! I've checked the formulas and the long expression seems fine. If you are finding a mistake after checking with a friend / textbook, check the calculations.

EDIT : Calculations are $\color{blue}{\mathrm{fine!}}$

Answer 2

Both of these are correct. Let $T_A$, $T_B$, $T_C$ be the events that the manuscript is typed by $A, B, C$ respectively. The conditional distribution of $X$ given $T_A$ is Poisson with parameter $2.6$, and similarly for $T_B$ and $T_C$ with parameters $3$ and $3.4$. Since $T_A$, $T_B$ and $T_C$ are mutually exclusive and exhaustive,

$$ \mathbb E[X] = \mathbb P(T_A) \mathbb E[X\mid T_A] + \mathbb P(T_B) \mathbb E[X\mid T_B] + \mathbb P(T_C) \mathbb E[X\mid T_C]$$

Similarly

$$ \mathbb P(X < 3) = \mathbb P(T_A) \mathbb P(X < 3 \mid T_A) + \mathbb P(T_B) \mathbb P(X < 3 \mid T_B) + \mathbb P(T_C) \mathbb P(X < 3 \mid T_C) $$