1

Then just from a brute force computation, I found that the largest value of $N$ needs to be 99 since any value greater than 99 is greater than 0.5. Is that correct? or is there another way to solve without brute force computation?

  • What is $L$? Where did you get the first equation? Note that the expected number of people there is $102$ so you should expect the theater to be close to full and $N$ should be larger. – Ross Millikan Oct 12 '22 at 15:41
  • sorry $L$ is the lambda value. – user8290579 Oct 12 '22 at 15:49
  • the total capacity of the theater is 100, how can the expected number of people be 100 – user8290579 Oct 12 '22 at 15:49
  • 5/min times 20 min. That does not depend on the capacity. If the capacity isn't large enough, some will be turned away – Ross Millikan Oct 12 '22 at 16:07
  • @RossMillikan oh okay, so what part should I adjust? – user8290579 Oct 12 '22 at 18:46
  • $\lambda = 100$ – Ross Millikan Oct 12 '22 at 18:59
  • @RossMillikan ah okay so I just redid the calculation in my original post and got $N\leq 99%$ Does that make sense? it seems like too high of a value for $N$. – user8290579 Oct 12 '22 at 23:36
  • As I said in my first comment the theater should be close to full. I didn't check the details, but that sounds right. For high $\lambda$ the Poisson distribution becomes normal, so you would expect the $50$th percentile to be just about the expected number. – Ross Millikan Oct 12 '22 at 23:53
  • @RossMillikan Ah okay thanks, I see what you're saying. I guess I was just confused because I think it means that the friend loses no matter what? or if N was that big, then the friend shouldn't make the bet because the probability of him winning is less than 0.5 for almost anything he can guess – user8290579 Oct 13 '22 at 14:45
  • I think that is correct. – Ross Millikan Oct 13 '22 at 14:46

0 Answers0