Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Probability town population

  1. Jun 29, 2009 #1
    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.
  2. jcsd
  3. Jun 29, 2009 #2
    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.
  4. Jun 29, 2009 #3
    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
  5. Jun 29, 2009 #4


    User Avatar
    Science Advisor
    Homework Helper

    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%.
  6. Jul 13, 2009 #5
    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)


    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: Jul 14, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook