I am new to this community :D
I have a question involving probability and probability distributions today. The question is as follows: a person typing a document has an error rate of 3 words per 1000 words. Given that this person typed a 5000 word document with 13 errors, what is the probability that there are 3 errors in the first 500 words?
It is immediately obvious, upon inspection, that this is a question suitable for a Poisson Distribution. According to the information given, I also know that $\lambda = 15$. The Poisson Distribution should be easy to calculate, but that's not the end here. I noticed the keyword "given", so this becomes a more challenging question where I need to calculate the conditional probability. How would I go about doing this? Thank you!
I think you can treat the number of errors in the first 500 words and the number of errors in the remaining 4500 words as two independent Poisson r.v.'s, and then calculate the conditional probability.
Let $X$ be the number of errors in the first $500$ words and $Y$ the number of errors in the remaining $4500$ words and let $X \sim Po(\lambda_1)$ and $Y \sim Po(\lambda_2)$ and assume they're independent.
It's straightforward to calculate the conditional probability...
$Pr(X=3|X+Y=13) = \frac{Pr(X=3) \; Pr(Y=13-3)}{Pr(X+Y=13)}$
$X+Y \sim Po(\lambda_x+\lambda_y)$, so $Pr(X=3|X+Y=13)$ simplifies to
$Pr(X=3|X+Y=13) = \binom{13}{3} \; (\frac{\lambda_1}{\lambda_1+\lambda_2})^3 \; (\frac{\lambda_2}{\lambda_1+\lambda_2})^{10}$
if the rate is considered per 1000 words then $\lambda_1 = 1.5$ and $\lambda_2=13.5$, and
$Pr(X=3|X+Y=13) = \binom{13}{3} \; (\frac{1.5}{15})^3 \; (\frac{13.5}{15})^{10}=0.0997$