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MathGroup Archive 2000

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Re: Setting up a probability problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg22177] Re: [mg22150] Setting up a probability problem
  • From: Daniel Lichtblau <danl at wolfram.com>
  • Date: Thu, 17 Feb 2000 01:24:10 -0500 (EST)
  • References: <200002160735.CAA18039@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Ron Lenk wrote:
> 
> I'm having trouble figuring out the mathematical formulation of this
> problem. I have n events, each of which has probability P (say 20%)
> of occurring at any given time. How large does n need to be in order
> to give me a confidence level c (say 90%) that I have m (say 5) events
> occurring simultaneously? Help in clarifying my thinking would be
> appreciated.


I'll change the language a bit to be more in accordance with standard
terminology (as best I remember it). We'll refer to n "trials" (or
experiments) for each of which we might have a "success." Regard time as
being fixed at some given instant, so it is not part of the formulation.

(i) What is the probability that exactly the first 5 trials are
successes and the rest not? It is (1/5)^5 * (4/5)^(n-5).

(ii) What is the probability that we have exactly 5 successes, not
necessarily from the first five trials? Just notice that there are
Binomial[n,5] ways to pick the 5 successful experiments from among the n
trials, and each has the same probability as in (i) above. So we get
Binomial[n,5]*(1/5)^5*(4/5)^(n-5)

(iii) What is the probability that we have at least 5 successes from n
trials? It is the sum of probabilities that we have exactly five
successes, exactly six successes, etc. (this is because these are all
mutually exclusive events).

Let us set up a function that is slightly more general, in that various
parameters are now given as variables rather than hard-wired constants.
We are given that the probability of a trial being successful is some
value, "ev." We want to find the probability of at least "min" successes
from among n trials. Then we have

probm[n_,ev_,min_] :=
	Sum[Binomial[n,j]*ev^j*(1-ev)^(n-j), {j,min,n}]

Last, we want to find the value of n for which this probability is equal
to 9/10 (more correctly, we want the smallest integer value of n for
which the probability meets or exceeds 9/10). We do this using a
root-finder.

In[79]:= probm[n_,ev_,min_] :=
        Sum[Binomial[n,j]*ev^j*(1-ev)^(n-j), {j,min,n}]         

In[80]:= FindRoot[probm[m,1/5,5]==9/10, {m,6}]
Out[80]= {m -> 37.8877}

This tells us we'll need 38 trials in order to guarantee a probability
of at least 90% that we obtain at least 5 successes.

To code this a bit more generally, we allow the threshhold probability
(your "confidence level" c) for our m successes to be given as a
parameter.

minTrials[conf_,ev_,min_] :=
  Ceiling[m /. First[FindRoot[probm[m,1/5,5]==conf, {m,6}]]]

In[84]:= minTrials[9/10,1/5,5]
Out[84]= 38


Daniel Lichtblau
Wolfram Research


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