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From |
Richard Williams <Richard.A.Williams.5@ND.edu> |

To |
statalist@hsphsun2.harvard.edu, <statalist@hsphsun2.harvard.edu> |

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:

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 isI 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)

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.

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Richard Williams, Notre Dame Dept of Sociology

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