1
$\begingroup$

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?

My answer:

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)?

$\endgroup$
1
$\begingroup$

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 $0.162$

$\endgroup$
1
  • $\begingroup$ Ahh yes thanks so much. I guess I was just going at it the wrong way. $\endgroup$
    – GilNorth
    Mar 18 '18 at 23:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.