1
$\begingroup$

Patients recover in a ward after various surgical procedures. There is a probability of recovery for each day of stay in the ward which depends upon which procedure was carried out and which surgeon carried out the procedure.

For example there is a probability 0.6 that patient A having had a hip replacement, carried out by surgeon B, will have recovered on day 4 of their stay in the ward.

Is it correct to assume that the calculation of the probability of 6 out of 10 free beds on day 1 could be treated as a sum of Binomial trials? Assuming that the recovery probabilities are independent.

P(6 beds free on day one) = P(beds 1,2,3,4,5,6 free)*p(beds 7,8,9,10 not free) + p(beds 2,3,4,5,6,7 free)*p(beds 1,8,9,10 not free + ...

Where P(beds 1,2,3,4,5,6 free) = P(bed 1 free)*p(bed 2 free)p(bed 3 free) ...

Where the probability of bed 1 being free depends on which day of recovery the patient is on, the type of surgical procedure that was carried out and which surgeon carried it out, assuming that these probabilities are known.

Are there any other approaches to calculating these probabilistic? For example using a probability generating function? A distribution as an approximation?

$\endgroup$
1
$\begingroup$

What you are interested in, generally, is a counting process. Length of stay for a particular patient is a function of several parameters: the type of surgery done but also their age among a whole host of other possible explanatory factors. Whether each bed is free is a Bernoulli trial: which takes values 0 or 1 with some probability, $p$. This is likely to vary from patient to patient. Laproscopy can be discharged within hours, but hip replacements may have a few days in the ER before being discharged to the SNF.

According to your description, these patient beds may independent, but the sum of independent Bernoulli trials does not make a single binomial trial. You also need identical distribution, that is that each $p$ must be the same. This is the assumption that all discharge probabilities are the same regardless of the procedure rendered. Sums of identically independent distributed Bernoulli trials takes the known binomial distribution.

However, in this day and age, even if the sum of beds is not binomial, it is relatively straightforward to use modeling and predict the number of beds that would be free at any particular time, regardless of whether they are identically distributed. The distribution of counts may "look" different than a binomial distribution, but calculating probabilities can be done easily, provided you have good knowledge about length of stay for the relatively common procedures.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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