0
$\begingroup$

Let Y be the number of calls to a particular hotline within 10 min. Suppose Y is a Poisson random variable with mean of 3. Find the probability that there are at most 4 calls given that there are already 2 calls within the 10 min.

I'm not really sure how to go about solving this. Am I supposed to be looking at this in terms of a binomial random variable? If not, then how would I set up the conditional probability?

$\endgroup$
0
$\begingroup$

By definition of conditional probability, \begin{align} \mathbb P(Y\leqslant 4\mid Y\geqslant 2) &= \mathbb P(Y=2\mid Y\geqslant 2) + \mathbb P(Y=3\mid Y\geqslant 2) + \mathbb P(Y=4\mid Y\geqslant 2)\\ &= \frac{\mathbb P(Y=2,Y\geqslant 2)}{\mathbb P(Y\geqslant 2)} +\frac{\mathbb P(Y=3,Y\geqslant 2)}{\mathbb P(Y\geqslant 2)} +\frac{\mathbb P(Y=4,Y\geqslant 2)}{\mathbb P(Y\geqslant 2)}\\ &= \frac{\mathbb P(Y=2)+\mathbb P(Y=3)+\mathbb P(Y=4)}{1 - (\mathbb P(Y=0) + \mathbb P(Y=1))}\\ &= \frac{e^{-3}(3^2/2! + 3^3/3! + 3^4/4!)}{1 - e^{-3}(1 + 3)}\\ &\approx 0.7693246. \end{align}

$\endgroup$
  • $\begingroup$ Why is it Y>=2 and not Y=2? $\endgroup$ – oamy Feb 18 at 2:08
  • $\begingroup$ From the wording of the question, it's given that there were at least two calls in the first $10$ minutes. $\endgroup$ – Math1000 Feb 18 at 3:12

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.