# Probability of red ball from countable many urns

So I have the following question and I just want to check if my reasoning is correct:

Consider countably many urns Uk, where urn Uk contains 1 red ball and 2k - 1 blue balls. We select an urn according to the Poisson distribution with parameter 4, i.e. the probability that we select urn Uk is P4({k}). We then draw a ball from this urn. What is the probability that we draw a red ball?

Let A = Chance of selecting Urn K = P4({k})

B|A = Chance of selecting a red ball given we selected Urn K = 1/k

From the equation of conditional probability we get that P(B|A) = P(A intersect B) / P(A)

Thus P(A intersect B) = P(B|A) P(A)

Should I be trying to find the probability of the intersection or should I be trying to find a different value for P(B|A)?

If we have a countably infinite number of urns, and given that we picked from urn $U_k$, see that we have $P(\text{red } | U_k) = \frac{1}{2k}$ because there is $1$ red ball and $2k-1$ blue balls in urn $U_k$, making it $2k$ in total.
We can then utilize the law of total probability to say: $$P(\text{red}) = \sum_{i=1}^\infty P(U_i) * P(\text{red }|U_i)$$ $$= \sum_{i=1}^\infty \frac{4^ie^{-4}}{2i(i!)}$$
WolframAlpha approximates this probability to be around $0.162$