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Re: st: R: probability question

From   Richard Williams <>
To, <>
Subject   Re: st: R: probability question
Date   Wed, 11 Jun 2008 16:01:58 -0500

At 03:28 AM 6/11/2008, Carlo Lazzaro wrote:
I need to calculate the probability of the following event:

That at least one patient out of 99 suffers from kidneycancer
(incidence 15/100000 pr year) over a time period of 15 years.

Dear Moleps,

searching the literature (Briggs A, Sculpher M, Claxton K. Decision
Modelling for Health Economic Evaluation. Oxford: Oxford University Press,
2006:51) I have found out a formula for converting rate into probabilities:



r= instantaneous rate, provided that it is constant over the period of
interest (t)
That might be more accurate than what I calculated before, but luckily it gives pretty much the same result. For one person, the probability of NOT getting cancer over 15 years is

p(0) = 1 - (1 - exp(-rt)) = exp(-rt) = exp(-15/100000 * 15), i.e.

. di exp(-15/100000 * 15)

The probability that all 99 people should be so lucky is

. di exp(-15/100000 * 15) ^ 99

Hence, the probability that at least one is not so lucky is

. di 1 - (exp(-15/100000 * 15) ^ 99)

Which is pretty close to my earlier calculation, which was

. di 1 - ((1 - 15/100000 )^15^99)

The difference is .00001337!

I imagine there might be some other complications in these calculations. The rate may not be constant across time; or people may die from something else before they get kidney cancer. But for the problem as stated, it sounds like there is a 20% chance of at least one person getting kidney cancer.

Richard Williams, Notre Dame Dept of Sociology
OFFICE: (574)631-6668, (574)631-6463
HOME: (574)289-5227
EMAIL: Richard.A.Williams.5@ND.Edu

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