Conditional probability in a Poisson distribution

Question

An industrial product can have two types of flaws. The number of flaws has a Poisson distribution of parameter $2$ for the first kind, and $4$ for the second. The number of flaws of one kind is independent from the other.

I need:

$a)$ Probability of a product having no flaws.

$b)$ Probability of a product having exactly one flaw.

$c)$ Given that a product is known to have exactly one flaw, what is the probability of it being a flaw of the 1st kind?

Here's what I've done:

X: Number of flaws of the 1st kind in a given product.

Y: Number of flaws of the 2nd kind in a given product.

$X \sim \operatorname{Poi}(2)$

$Y \sim \operatorname{Poi}(4)$

So:

$\mathbb{P}(X = x) = \frac{2^xe^{-2}}{x!}$

$\mathbb{P}(Y = y) = \frac{4^ye^{-4}}{y!}$

Since both are independent: $\mathbb{P}(X = x, Y = y) = \frac{2^x4^ye^{-6}}{x!y!}$

$a)$ $\mathbb{P}(X = 0, Y = 0) = \frac{2^04^0e^{-6}}{0!0!} = e^{-6}$

$b)$ $\mathbb{P}(X = 1, Y = 0) + \mathbb{P}(X = 0, Y = 1) =\frac{2^14^0e^{-6}}{1!0!}+\frac{2^04^1e^{-6}}{0!1!} = 6e^{-6}$

I'm having trouble with $(c)$. I think I have to use $\mathbb{P}(X = x | A) = \frac{\mathbb{P} (X = x)}{\mathbb{P}(A)} \mathbb{I} \{X \in A\}$ where $A$ is "the product has at least one flaw" and $\mathbb{P}(A) = 6e^{-6}$.

I'm not sure what I should put in the indicator function, or if it has any bearing on the final result. I think after I figure that out, the answer should be $\mathbb{P}(X = 1 \mid A)$.

Thanks.

Answer 1

Using the definition of conditional probability \begin{equation} \mathbb{P}(1^{st} \text{ kind} \vert \text{One Flaw}) = \frac{\mathbb{P}(1^{st} \text{ kind} \cap \text{One Flaw}) }{\mathbb{P}(\text{One Flaw})} \end{equation} that is according to your writing and notations: \begin{equation} \mathbb{P}(1^{st} \text{ kind} \vert \text{One Flaw}) = \frac{\mathbb{P}(X = 1, Y = 0)}{\mathbb{P}(X = 1, Y = 0)+\mathbb{P}(X = 0, Y = 1)} \end{equation} You get according to your math \begin{equation} \mathbb{P}(1^{st} \text{ kind} \vert \text{One Flaw}) = \frac{\frac{2^14^0e^{-6}}{1!0!}}{6e^{-6}} = \frac{2}{6}=\frac{1}{3} \end{equation}