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Patients enter emergency department following a Poisson process with intensity 6 patients per hour.

  1. Calculate the probability that during 90 minute interval there will have entered exactly 7 patients.
  2. The last patient arrived at 13:00. What is the probability that the next patient will arrive before 13:15?

So, my solution:

  1. $$\lambda = 1.5 * 6 = 9 \\ P(X = 7) = \frac{e^{-9} 9^7}{7!} \approx 0.117$$

  2. $$\ \ \lambda = \frac{6}{60} \cdot 15 = \frac{3}{2} \\ P(X = 1) = \frac{e^{-3/2}\cdot \frac{3}{2}}{1!} \approx 0.552$$

I am not sure about the second one.

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  • $\begingroup$ (1) In R, where dpois is a Poisson PDF, code dpois(7,9) returns $0.1171161.$ $\endgroup$
    – BruceET
    Commented Jan 12, 2021 at 8:23

1 Answer 1

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(2) The waiting time for the next event is $W \sim \mathsf{Exp}(\mathrm{rate\,}=\lambda = 6.)$ You seek $P(W \le 3/4) = 0.9889.$

One expects six events per hour (one every 10 min.) on average, so it is very likely the next event occurs within 45 min.

In R, where pexp denotes an exponential CDF, and using R as a calculator:

pexp(3/4, 6)
[1] 0.988891
1 - exp(-6*(3/4))
[1] 0.988891

enter image description here

curve(dexp(x,6), 0, 1.5, lwd=2, ylab="PDF", xlab="w", 
      main="Density of EXP(6)") 
 abline(v=0, col="green2");  abline(h=0, col="green2")
 abline(v = 3/4, col="red", lty="dotted", lwd=2)
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  • $\begingroup$ Sorry, it was a typo, I meant 15 minutes. So with dpois(1, 3/2) = 0.334, but calculating it the way I did it with the formula, gives me a result of 0.552. The 1-exp(-1/4*6) gives a result of 0.776. I understand that dpois and 1-exp are linked. But the previous commenter accepted my first answer as correct and the function dpois gives the result I computed. I don't understand why now I should use the 1-exp function. Why my own computed result is incorrect? $\endgroup$
    – Gianni D'Adova
    Commented Jan 12, 2021 at 12:11
  • $\begingroup$ This is getting to be a mess: As I said in a previous comment, your answer to (1) is OK. For (2) if you've now decided you want $P(W < 1/4).$ then it's pexp(1/4,6), which returns $0.7768698.$ Best to use $W$ for waiting times. But if you insist on using Poisson, then you need to use $Y\sim\mathsf{Pois}(3/2)$ and find $P(Y\ge1) = 1-P(Y=0),$ which computes in R as 1 - dpois(0, 3/2) returning $0.7768698.$ So the method and answer you show for (2) are wrong. $\endgroup$
    – BruceET
    Commented Jan 12, 2021 at 18:11
  • $\begingroup$ Ok, thank you. One more question. If you calculate the probability that during the 15 min. interval there will be another event by 1-dpois(0,3/2), does that imply that you calculate the probability that during these 15 min. there will occur 1 or more event? $\endgroup$
    – Gianni D'Adova
    Commented Jan 12, 2021 at 19:34
  • $\begingroup$ The event that the next patient arrives before 13;15 obviously allows the possibility that one or more will arrive before 13:15. // That's the reason I said it's best (= least confusing) to use exponential dist'n for waiting times. // If you want the probability that exactly one patient arrives btw 13:00 and 13:15, then you need to say so. // This site works best with exactly one clearly stated question at a time. So I'm signing off now, before we get chastised for 'chatting' in comments. $\endgroup$
    – BruceET
    Commented Jan 12, 2021 at 21:20
  • $\begingroup$ No, I wanted exactly what you answered for. Thanks. $\endgroup$
    – Gianni D'Adova
    Commented Jan 12, 2021 at 23:08

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