Prob 33, Ch. 3 - Two proofreaders reading a book with n typos - Discrete probability distributions, Blitzstein and Hwang I would like someone to verify my result to parts (a) and (b) of this problem.


  
*A book has $n$ typos. Two proofreaders Prue and Frida independently read the book. Prue catches each typo with probability $p_{1}$ and misses it with probability $q_{1}=1-p_{1}$, and likewise for Frida, who has probabilities $p_{2}$ if catching and $q_{2}=1-p_{2}$ of missing each typo. 
  
  
  Let $X_{1}$ be the number of typos caught by Prue, $X_{2}$ be the number of typos caught by Frida and $X$, be the number of typos caught by atleast one of the proofreaders.
(a) Find the distribution of $X$.

Solution.
Define an r.v. $I_{p}$.
$$\begin{align}
I_{p}&=1 \text{, if Prue catches an error}\\
&=0 \text{, if Prue misses to catch an error}
\end{align}$$
We have,
$$\begin{align}
P(I_{p}=1)&=p_{1}\\
P(I_{p}=0)&=1-p_{1}
\end{align}$$
On similar lines,
$$\begin{align}
P(I_{f}=1)&=p_{2}\\
P(I_{f}=0)&=1-p_{2}
\end{align}$$
We have,
$\begin{align}
P(I_{p}=1\cup{I_{f}}=1)&=P(I_{p}=1)+P(I_{f}=1)-P(I_{p}=1,I_{f}=1)\\
&= p_{1}+p_{2}-p_{1}p_{2}
\end{align}$
Thus, define success as, a typo is caught by atleast one of the proof-readers. Each of the $n$ typos can be caught or missed by atleast one proofreader. These can be conceived as $n$ independent Bernoulli trials. Therefore, $X\sim{Binomial(n,p_{1}+p_{2}-p_{1}p_{2})}$.
$\displaystyle{P(X=x)={{n}\choose{x}}(p_{1}+p_{2}-p_{1}p_{2})^{x}(1-p_{1}-p_{2}+p_{1}p_{2})^{n-x}}$

(b) For this part only, assume $p_{1}=p_{2}$. Find the conditional distribution of $X_{1}$, given that $X_{1}+X_{2}=t$.

Solution.
$\begin{align}
P(X_{1}=x|X_{1}+X_{2}=t)&=\frac{P(X_{1}=x,X_{1}+X_{2}=t)}{P(X_{1}+X_{2}=t)}\\
&=\frac{{{n}\choose{x}}p^{x}(1-p)^{n-x}{{n}\choose{t-x}}p^{t-x}(1-p)^{n-(t-x)}}{{{2n}\choose{t}}p^{t}(1-p)^{2n-t}}\\
&=\frac{{{n}\choose{x}}{{n}\choose{t-x}}}{{{2n}\choose{t}}}
\end{align}$
 A: Looks good to me. I think you may want to add
$\begin{align}
P(X_{1}=x|X_{1}+X_{2}=t)&=\frac{P(X_{1}=x,X_{1}+X_{2}=t)}{P(X_{1}+X_{2}=t)}\\
&=\frac{P(X_{1}=x,x+X_{2}=t)}{P(X_{1}+X_{2}=t)}\\
&=\frac{P(X_{1}=x)P(x+X_{2}=t)}{P(X_{1}+X_{2}=t)}\tag{*}\\
&=\frac{P(X_{1}=x)P(X_{2}=t-x)}{P(X_{1}+X_{2}=t)}\\
&=\frac{{{n}\choose{x}}p^{x}(1-p)^{n-x}{{n}\choose{t-x}}p^{t-x}(1-p)^{n-(t-x)}}{{{2n}\choose{t}}p^{t}(1-p)^{2n-t}} \tag{**}\\
&=\frac{{{n}\choose{x}}{{n}\choose{t-x}}}{{{2n}\choose{t}}}
\end{align}$
$(*)$ $X_1$ and $X_2$ are independent
$(**)$ This assumes $X_1$ and $X_2$ are binomial. Now, a binomial random variable is the sum of independent Bernoulli random variables. Are the Bernoulli random variables that make up each independent as well i.e. are the typo catching's for each person independent?
I mean
$$X_1 = B_1 + ... B_n$$
$$X_2 = C_1 + ... C_n$$
I guess $B_i$ and $C_j$ are independent for any i,j = 1, 2, ..., n, not necessarily distinct from the fact that $X_1$ and $X_2$ are independent, but are $B_1, B_2, ..., B_n$ independent? What about $C_1, C_2, ... C_n$?
