# Re: st: R: probability question

 From Svend Juul To Subject Re: st: R: probability question Date Thu, 12 Jun 2008 08:51:18 +0200

```Moleps asked, and Carlo and Richard commented:

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.

====================

Carlo:

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)

======================

Richard:

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.

======================================================================

To get the expected proportion over 15 years, 15*15/100000 = 0.00225
is almost right with a rare event, but as Carlo pointed out, the
correct conversion from a rate to a proportion says:

. display 1-exp(-15/100000 * 15)
.00224747

Now, the official -bitest- command does the rest of the job:

. bitesti 99 1 0.00224747

N   Observed k   Expected k   Assumed p   Observed p
------------------------------------------------------------
99          1     .2224995       0.00225      0.01010

Pr(k >= 1) = 0.199685  (one-sided test)
Pr(k <= 1) = 0.978786  (one-sided test)
Pr(k >= 1) = 0.199685  (two-sided test)

note: lower tail of two-sided p-value is empty

Pr(k >= 1) = 0.199685

Svend

__________________________________________
1
Svend Juul
Institut for Folkesundhed, Afdeling for Epidemiologi
(Institute of Public Health, Department of Epidemiology)
Vennelyst Boulevard 6
DK-8000  Aarhus C, Denmark
Phone:  +45 8942 6090
Home:   +45 8693 7796
Email:  sj@soci.au.dk
__________________________________________

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