Catching Errors While Proofreading a Document

A document has n typos, and there are two proofreaders working independently on proofreading the document. Proofreader A catches typos with probability $p_{1}$ and misses them with probability $q_{1} = 1 - p_{1}$. Proofreader B catches typos with probability $ p_{2}$ and misses them with probability, you guessed it, $q_{2} = 1 - p_{2}$. Let $X_{1}$ be the number of typos caught by Proofreader A, and $X_{2}$ be the number of typos caught by Proofreader B, and $X$ be the number of typos caught by at least one of them.

(a) What is the distribution of X?

Solution: The easiest way to think about this is to think of the case where neither of them catch a typo. The probability of both of them missing a typo, (each Proofreader's miss probability is independent of the other's), is gotten through multiplication, $q_{1}q_{2}$. So, the probability of at least one of them catching that typo is $1 - q_{1}q_{2}$. Thus, the distribution is
$X\sim Bin(n,1-q_{1}q_{2})$

(b) Assume $p_{1} = p_{2}$, what is the conditional distribution of  $X_{1}$ given that $X_{1}+X_{2} = t$?

Solution: Using Bayes' Rule, we have:

$P(X_{1}=k|T=t) = \frac{P(T=t|X_{1}=k)P(X_{1}=k)}{P(T=t)}$


with $T\sim Bin(2n,p)$

$P(X_{1}=k|T=t) = \frac{\binom{n}{t-k}p^{t-k}q^{n-t+k}\binom{n}{k}p^{k}q^{n-k}}{\binom{2n}{t}p^{t}q^{2n-t}} = \frac{\binom{n}{t-k}\binom{n}{k}}{\binom{2n}{t}}$

$X_{1}\sim HGeom(n,n,t)$

Notes: $P(T=t|X_{1}=k)$ considers the probability of T = t given $X_{1}=k$, which means we only have to consider how $X_{2}$ affects the calculation. t = total catches by Proofreader A + total catches by Proofreader B. If A catches k typos, then B catches t - k typos.