Over a long period of time ,a plumber finds that on average he receives 2 emergency calls per week .Work out the probability of one emergency call on one day assuming the plumber is available for emergency calls five days a week .

I am getting 0.268 while answer is 0.938 . Please help !?

  • $\begingroup$ maybe at least one? $\endgroup$ – quester Oct 20 at 7:39
  • 2
    $\begingroup$ At least once gives a probability of about $0.33$. Your answer appears to be right unless you forgot to mention any details in the problem. It's clear that you got your answer by finding $\mathsf P(X=1)$ where $X\sim\mathsf{Pois}\left(\frac{2}{5}\right)$. However, in the future, please show how you got your answer. $\endgroup$ – Remy Oct 20 at 7:41
  • $\begingroup$ Thanks ,yes will show working in the future ! $\endgroup$ – Sara Oct 20 at 7:48
  • $\begingroup$ You should still edit to show how you got 0.268 $\endgroup$ – Glen_b -Reinstate Monica Oct 20 at 11:39

If $X_5 \sim \mathsf{Pois}(\lambda_5 = 2),$ and assuming that emergency calls tend to be spread evenly among days, then $X_1 \sim \mathsf{Pois}(\lambda_1 = 2/5)$ describes the distribution of emergency calls on any one particular day.

Then using the Poisson PDF function dpois in R, we get $P(X_1 = 1) =0.2681$ and (as @remy says) $P(X_1 \ge 1) = 1 - P(X_1 = 0) = 0.3297.$

 dpois(1, 2/5)
 [1] 0.268128

 1 - dpois(0, 2/5)
 [1] 0.32968

Here is a bar graph (made with R) of the PDF of $\mathsf{Pois}(2/5).$

enter image description here

x=0:5;  pdf=dpois(x, 2/5)
plot(x, pdf, type="h", lwd=3, ylab="PDF", xlab="x", 
    ylim=c(0,max(pdf)), main="PDF of POIS(2/5)")
  abline(v=0, col="green2");  abline(h=0, col="green2")

Note: I've tried finding answers to several plausible (and some not quite so plausible) alternative questions. It may be worthwhile to notice that $P(X_1 \le 1) = 0.3984.$ That would be the answer to "at most 1 emergency call on a particular day."

ppois(1, 2/5)
[1] 0.9384481
  • $\begingroup$ So the question should have been atmost 1 . $\endgroup$ – Sara Oct 21 at 6:15
  • $\begingroup$ Not 'almost', but 'at most'. Not more than 1. $P(X_1 \le 1)$. Question as you have it makes sense. But 'at most 1' gets you the answer claimed. $\endgroup$ – BruceET Oct 21 at 6:58

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