# 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 $0.01$

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.

• 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. Nov 10, 2019 at 18:29

$$X$$ 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.

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

• Thank you for the answer! Nov 10, 2019 at 18:56