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Re: st: R: probability question
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:
p=1-exp(-rt)
where:
p=probability;
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)
.99775253
The probability that all 99 people should be so lucky is
. di exp(-15/100000 * 15) ^ 99
.8003149
Hence, the probability that at least one is not so lucky is
. di 1 - (exp(-15/100000 * 15) ^ 99)
.1996851
Which is pretty close to my earlier calculation, which was
. di 1 - ((1 - 15/100000 )^15^99)
.19969847
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: [email protected]
WWW: http://www.nd.edu/~rwilliam
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