At a certain men's college, the probability that a student selected at random on a given day will require a hospital bed is $\dfrac {1}{5000}.$ If there are $8000$ students, how many beds should the hospital have so that the probability that a student will be turned away for lack of bed is less than $0.01 $

Attempt: Let $X$ be the random variable which denotes the number of students requiring a hospital bed on a given day.

Trying to use the poisson distribution, we obtain, $\lambda = 8000 \cdot \dfrac {1}{5000} = \dfrac {8}{5}$. Then, using poisson distribution : $P(X=x) = \dfrac {e^{-\lambda}\cdot \lambda^x}{x!}$

Let $n$ be the required number of beds ( $n < 8000)$. Then, by the given requirement, the probability that $n$ beds will be occupied should be less than $0.01 \implies \dfrac {e^{-\frac{8}{5}}\cdot \big( \frac{8}{5}\big)^n}{n!} \le 0.01$

$\implies \Big ( \dfrac{8}{5} \Big)^n \cdot \dfrac{1}{n!} \le 0.01 \cdot e^{\frac{8}{5}}$

Solving for which we get $n \ge 6$.

Is this the correct approach? Thanks a lot for your help.

  • 1
    $\begingroup$ A Poisson approximation will work. As @gunes says, your mistake is to use the PDF not the CDF. In R, code x=1:10; pr = ppois(x, 8/5); min(x[pr >= .99]) returns $5.$ Also, using binomial, x=1:10; pr = pbinom(x, 8000, 1/5000); min(x[pr >= .99]) gives the same result. $\endgroup$
    – BruceET
    Nov 10, 2019 at 18:29

2 Answers 2


$X$ is binomial and we try to find the minimum $n$ that satisfy the following: $P(X>n)\leq 0.01$, i.e. number of ill students is larger than number of beds in the hospital. This is equivalent to $P(X\leq n)\geq 0.99$. $$P(X\leq n)=\sum_{i=0}^n P(X=i)=\sum_{i=0}^n {8000\choose i}\left(1\over 5000\right)^i\left(4999\over 5000\right)^{8000-i}$$ This is numerically challenging (still can be done using softwares like Matlab). A better way is using Poisson approximation since $n$ is large and $p$ is small. The rate parameter $\lambda$ is defined as $np=8/5$. Substituting into the summation above: $$P(X\leq n)=\sum_{i=0}^n P(X=i)=\sum_{i=0}^n \frac{e^{-\lambda}\lambda^i}{i!}=e^{-\lambda}\left(1+\lambda+\frac{\lambda^2}{2}+\frac{\lambda^3}{6}...+\frac{\lambda^n}{n!}\right)$$

A couple trials will reveal that you actually need $n=5$ beds at minimum. Your attempt lacks the CDF approach.

  • $\begingroup$ Thank you for the answer. Does that mean, a CDF approach refines the answer obtained using the pdf? But essentially are both approach correct? $\endgroup$
    – MathMan
    Nov 10, 2019 at 18:44
  • 3
    $\begingroup$ @MathMan PDF (or PMF) is not enough because you find $P(X=n)$, however you actually need $P(X\leq n)$. $\endgroup$
    – gunes
    Nov 10, 2019 at 18:47
  • $\begingroup$ Got it. Thanks a lot $\endgroup$
    – MathMan
    Nov 10, 2019 at 18:55

Comment continued: In case the 'search' implied by the brackets [ ] in R is not clear, and in order to show how well the Poisson approximation to the binomial distribution works in this example, here is a table of relevant Poisson and binomial CDFs.

x = 1:10; p.pr = ppois(x, 8/5); b.pr = pbinom(x,8000,1/5000)
cbind(x, p.pr, b.pr)
      x      p.pr      b.pr
 [1,]  1 0.5249309 0.5249116
 [2,]  2 0.7833585 0.7833688
 [3,]  3 0.9211865 0.9212058
 [4,]  4 0.9763177 0.9763310
 [5,]  5 0.9939597 0.9939657  # first entry exceeding 0.99
 [6,]  6 0.9986642 0.9986663
 [7,]  7 0.9997396 0.9997401
 [8,]  8 0.9999546 0.9999548
 [9,]  9 0.9999929 0.9999929
[10,] 10 0.9999990 0.9999990
  • $\begingroup$ Thank you for the answer! $\endgroup$
    – MathMan
    Nov 10, 2019 at 18:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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