Calculating the conditional probability of two poisson distributions where one parameter is unknown.

A production line creates products at a 60% success rate (40% are defective). The number of object produced in an hour is a RV Y which follows a poisson distribution of parameter 20. A RV X represents the number of defective items produced in an hour. Calculate the probability that X=k knowing that Y=n.

This is a homework assignment. I am looking for someone to point me in the right direction as I have no clue how to start.

Since the rate of defection is $\frac{2}{5}$, then if you produce $m$ items in an hour, the expected number of defect items is $\frac{2}{5}m$. Hence $X|y\sim\mathcal{P}\left(\frac{2}{5}y\right)$, and $\Pr(X=k|Y=n)=\frac{2^k n^k e^{-\frac{2}{5}n}}{5^k k!}$
• you said that $40\percent$ – Rizky Reza Fujisaki Oct 17 '16 at 6:40
• i am sorry, okay, you said that 40% (or $\frac{2}{5}$) of the products are defective, that is why the that is the rate of defection. Imagine you have 5 products, then you expect there are only 3 products in good shape. In general, knowing you produced $Y=n$ products, then the expected ($\mu$ value) number for defective products is $\frac{2}{5}n$, and the last step is just submit the number to the formula of poisson density function – Rizky Reza Fujisaki Oct 17 '16 at 6:46