What is the Probability of 5 People in a Small Town Getting Cancer in One Year?

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Discussion Overview

The discussion revolves around calculating the probability of exactly 5 people in a small town of 13,700 getting cancer in one year, given a larger population of 70,000,000 with an annual incidence of cancer between 3,000 to 5,000 cases. The focus includes probabilistic modeling and the application of statistical formulas.

Discussion Character

  • Mathematical reasoning
  • Exploratory
  • Technical explanation

Main Points Raised

  • One participant proposes a formula involving the ratio of cancer cases to the total population and the town's population, but expresses uncertainty about the correct approach.
  • Another participant challenges the initial approach, suggesting that it does not accurately represent the probability of selecting 5 cancer cases from the town.
  • A later reply reiterates the need to find the probability of randomly selecting 3,000 to 5,000 people from the larger population, with exactly 5 cases from the town.
  • One participant suggests a formula involving the probability of success and failure, using the binomial distribution to calculate the likelihood of exactly 5 cases in the town.
  • Another participant mentions the use of factorials in the calculations, specifically referencing the binomial coefficient.

Areas of Agreement / Disagreement

Participants express differing views on the correct method for calculating the probability, with no consensus reached on a single approach. Some participants agree on the use of binomial distribution, while others question the initial assumptions and calculations.

Contextual Notes

Participants note the importance of defining the probability of success and failure, and the implications of the population size on the calculations. There is also mention of the need for factorials in the probability calculations, indicating a reliance on combinatorial mathematics.

csnsc14320
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Feeling a bit rusty on my probability -

Say you have 70,000,000 people in a country and a small town with 13,700 people.

Now, say 3,000 - 5,000 people of the 70,000,000 people become infected with cancer per year.

What are the chances 5 people from the small town get infected in one year?

This isn't for a class so I welcome the right answer.

I was thinking something along the lines of

(3,000/70,000,000 * 13,700/70,000,000)^5 percent chance to(5,000/70,000,000 * 13,700/70,000,000)^5 percent chance, but I'm can't recall at the moment.
 
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I don't think that this is the right way.

This looks like the probability of randomly choosing five people in the country and picking all five in the town and all five have cancer.

I remember doing some problems like these and they use factorials in them. I'm trying to think of the exact equation but it doesn't come to mind.
 
bucher said:
I don't think that this is the right way.

This looks like the probability of randomly choosing five people in the country and picking all five in the town and all five have cancer.

I remember doing some problems like these and they use factorials in them. I'm trying to think of the exact equation but it doesn't come to mind.

Yeah, I guess in words what I am trying to do is find the probability of randomly choosing 3000-5000 people from 70,000,000 that exactly 5 of them would be from my town of 13,700 people
 
csnsc14320 said:
Yeah, I guess in words what I am trying to do is find the probability of randomly choosing 3000-5000 people from 70,000,000 that exactly 5 of them would be from my town of 13,700 people

Let's say that every person in your country has a p = 1 - q = 4000/70 000 000 chance of getting cancer. Then the chance that exactly k people get cancer in a town of 13,700 is p^k q^(13 700 - k) (13 700 choose k), which is small (0.11%) for k = 5.

But then again if there are 2000 such villages, the chance that at least one has 5 or more cases of cancer is 92%.
 
Nevermind, CRGreathouse got it. (p=0.0011195)

For csnc: the factorials come from the (n choose x) part of the formula. That would be defined as n! / (x!*(n-x)!) but nCr function on calculator works great. ;)
That substitutes into

p(x)=(n choose x) * p^x * q^(n-x)

,where

p=probability of "success" (sick person) =4000/70,000,000 (assuming we reduce the range to a single value and distribution is approximately symmetric),
q=1-p=probability of "failure" (well person),
n=number of trials (13,700), and
x=# of successes in n trials (looking for 5)
 
Last edited:

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