# Calculate $\mathbb{P}[Y=y|X=x]$ where $X$=# claims reported diring firs year, $Y$=# total of claims that will eventually be reported

A property-casualty insurance company issues automobile policies on a calendar year basis only. Let $$X$$ be a random variable representing the number of accident claims reported during calendar year 2005 on policies issued during calendar year 2005. Let $$Y$$ be a random variable representing the total number of accident claims that will eventually be reported on policies issued during calendar year 2005. The probability that an individual accident claim on a 2005 policy is reported during calendar year 2005 is $$d$$. Assume that the reporting times of individual claims are mutually independent. Assume also that $$Y$$ has the negative binomial distribution, with fixed parameters $$r$$ and $$p$$, given by $$\mathbb{P}[Y=y]=\binom{r+y-1}{y}p^{r}(1-p)^{y}$$ for $$y=0,1,\ldots$$. Calculate $$\mathbb{P}[Y=y|X=x]$$ the probability that the total number of claims reported on 2005 policies is $$y$$, given that $$x$$ claims have been reported by the end of the calendar year.

Remark: I know that the solution requires the use of Baye's Theorem and Theorem of Total Probability, and the identity $$\binom{y}{x}\binom{r+y-1}{y}=\binom{r+x-1}{x}\binom{(r+x)+(y-x)-1}{y-x}$$.

I have not been able to correctly describe $$X$$ or include $$d$$ in the analysis. I need your help to understand this better.

$$Y$$ represents the total random claim count on year 2005 issued policies. $$X$$ represents the subset of those claims that were reported in the same calendar year of issue. For a sufficiently large number of claims made, you would expect that $$X \approx Y \cdot d$$. Another way to say this is that given that $$Y$$ claims are reported across year 2005 issued policies, the random number $$X$$ of these claims that were reported in the same year is binomial, since each such claim is independent and identically distributed with probability $$d$$ of occurring in the same calendar year. Formally, we would write $$X \mid Y \sim \operatorname{Binomial}(Y, d), \\ \Pr[X = x \mid Y = y] = \binom{y}{x} d^x (1-d)^{y-x}, \quad x \in \{0, 1, \ldots, y\}.$$ So now we apply the law of total probability to obtain the marignal (unconditional) distribution of $$X$$: $$\Pr[X = x] = \sum_{y=0}^{\infty} \Pr[X = x \mid Y = y] \Pr[Y = y].$$ Then, we apply Bayes' theorem: $$\Pr[Y = y \mid X = x] = \frac{\Pr[X = x \mid Y = y]\Pr[Y = y]}{\Pr[X = x]},$$ where the numerator is simply the summand in the previous equation, and the denominator is the value of that sum.
I have left the actual computation of these to you as an exercise, as I regard the salient aspect of your question to be the fact that $$X \mid Y$$ is binomial.