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Correct Combination Thought Logic Being Followed

  1. Sep 30, 2008 #1
    For fun this semester I decided to take a probability class and doing a lot of the random problems from the book to learn the material. I am finding that sometimes I am over complicating some of these problems.

    Here is a problem I came across. When I punched in the numbers it didn't seemed valid to me. Unfortunately, the book being used doesn't have odd or even answers in the back of the book.

    The problem says.

    "Two new drugs are to be tested using a group of 60 laboratory mice, each tagged with a number for identification purposes. Drug A is to be given to 22 mice, drug B is to be given to another 22 mice, and the remaining 16 mice are to be used as controls. How many ways can the assignment of treatments to mice be made? A single assignment involves specifying the treatment for each mouse -- whether drug A, drug B or no drug."


    My logic.

    I asked myself how many different ways can drug A be taken? I say that 22 mice have to be given it and there are 60 mice total. To mean, that means taking n! / r!(n-r)!

    n is 60
    r is 22

    Using these values I get this

    14,154,280,149,473,100

    Seeing that number sent red flags here.


    Is my logic faulty here? I think it is but looking for some reassurance from others. Thanks.
     
  2. jcsd
  3. Sep 30, 2008 #2

    CRGreathouse

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    That looks fine. Now how many ways can the remaining mice be assigned?
     
    Last edited: Sep 30, 2008
  4. Sep 30, 2008 #3
    Really, this is fine? If it is that number just amazes me.

    CRGreathouse,

    Using my logic here is my response.

    There are 14,154,280,149,473,100 ways for 22 unique mice to use drugA.
    There are 14,154,280,149,473,99 ways for 22 unique mice to use drugB.
    There are 1.49608 X 10^14 ways for 16 unique mice to use no drugs.

    Given this I need to add up all these numbers.

    However, these numbers still don't seem right to me. I am still working on this problem.
     
  5. Sep 30, 2008 #4

    CRGreathouse

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    Your answer is not correct. You first choose 22 of 60 mice to assign to drug A... good. You then assign any of the 60 to drug B... not so good. Some mice are being assigned two drugs!


    As far as the largeness of the numbers is concerned, you can think of things in terms of the information needed to specify the assignment. A bit can store 2 different values. If there were 2 groups that each mouse could be assigned to (and you could have any number of mice in either group) then it would take 60 bits to specify which assignment was chosen since there are 2^60 ways to assign them. In your case it takes lg(14,154,280,149,473,100) = 53.6... bits to specify which 22 of 60 mice are assigned to drug A. So if 14,154,280,149,473,100 seems large, instead think about 54 bits (7 bytes). Does that seem reasonable?
     
  6. Sep 30, 2008 #5
    Oh, I see now.

    - 22 mice for drug A
    - then when i do drug B I can only use 38 (for n) to prevent double counting
    - then when i do no drugs I have to use 18 (for n)

    Yes, you bit example seems very reasonable. Thanks.
     
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