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### Post-Operative Complications

```Date: 04/25/2002 at 00:48:54
From: Jay Smith
Subject: Statistics

Q. The rate of post-operative complication in a complex vascular
reconstruction procedure, ranging in severity from wound separation
or infection to death, is 12%.

(i) For 40 such procedures, what are the chances that the number of
patients who suffer post-operative complications will be more than
ten? Less than expected?

(ii) In a series of 65 such procedures, determine the most probable
number of patients who suffer post-operative complications. What is
the probability?
```

```
Date: 04/25/2002 at 10:55:57
From: Doctor Mitteldorf
Subject: Re: Statistics

The key to solving problems of this kind is the binomial distribution
formula:  C(n,r)*(p^r)*(1-p)^(n-r)

This represents the probability that you will have r occurrences out
of a sample of n, when the probability of each occurrence separately
is p. The factor p^r represents probabilties for those r occurrences.
The factor (1-p)^(n-r) is the corresponding probability factor for the
remaining (n-r) instances where the complication does not occur.
C(n,r) is the number of ways this can happen, equal to the number of
ways to select r objects out of n total:

n!
--------------
(n-r)! r!

For example, in your question, p = 0.12 and 1-p = 0.88. In part (i)
you are talking about n = 40. The probability for more than 10
complications will be the sum of the binomial formula for r = 11
through 40.  (You may find that if you calculate the first few terms
after r = 11 that they are decreasing rapidly with r, and you can
neglect most of these to a good approximation.)

When they ask about "less than expected," they are referring to the
expected number of occurrences out of 40, which is just 12% of 40 =
4.8. Thus "less than expected" is the sum for r = 0 through r = 4.

- Doctor Mitteldorf, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Probability

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