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Combinations (n choose r)

  1. May 27, 2005 #1
    Hi,

    I just started a course about Statistics and I was wondering if I am doing the following question right.

    Q: A student enters 2 of 6 positive numbers with "-" signs and 3 of 4 negative numbers with "+" signs. If at some stage, a program chooses 3 distinct numbers from these 10 random numbers without replacement and multiplies them, we want to calculate the probability that at this stage no error occurs.

    Totally, how many different ways are there of choosing three numbers for the program at that stage?

    My answer:

    So, just say I entered these 10 numbers

    -1 -2 3 4 5 6 7 8 9 -1, but I really intended for 1 2 3 4 5 6 -7 -8 -9 -1. Therefore, there are in total 5 numbers that are entered incorrectly. In order for no error, the computer must choose 3 numbers from a possible 5 numbers that were entered correctly. Therefore, the number of ways would be

    5 choose 3 = 5! / (3! (5 - 3)!) = 10 ways correct?

    Thanks for your time.
     
  2. jcsd
  3. May 27, 2005 #2

    honestrosewater

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    What happens if you choose exactly 2 of the incorrectly labelled numbers?
    Does a * b = -a * -b and -a * b = a * -b?
     
  4. May 27, 2005 #3
    If you choose two incorrect numbers, then it will result in an error, which we do not want.
     
  5. May 27, 2005 #4

    honestrosewater

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    Say you choose -1, -2, and 3, but -1 and -2 should have been 1 and 2.
    -1 * -2 * 3 does not equal 1 * 2 * 3?
     
  6. May 27, 2005 #5
    Oh, I see what you mean. Well, technically that is true. Hmm, I guess we need to take that into consideration when doing the question. I'm stumped now lol
     
  7. May 27, 2005 #6
    A very naive approach would be to just consider different cases of the three numbers being chosen.
    Case 1 : all three are correct numbers
    Case 2 : two are correct and one is the wrong number
    Case 3 : one correct and two wrong
    Case 4 : all three are wrong

    Now for each case, calculate the probability and see where u can go from there?

    -- AI
     
  8. May 27, 2005 #7
    My answer of 10 is incorrect then? Where I just omit 5 numbers of the already given 10? 5 correct numbers, choose 3. I don't quite understand what you mean.
     
  9. May 27, 2005 #8
    Katakonik,
    that answer just gives the answer for case 1.
    what abt the other cases?

    -- AI
     
  10. May 27, 2005 #9

    honestrosewater

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    You want to know the probability of no error occuring. So figure out when errors occur:
    1) Exactly 0 incorrectly labelled numbers are chosen. This obviously results in no error.
    2) Exactly 1 incorrectly labelled number is chosen. Does this result in error?
    3) Exactly 2 ...
    4) Exactly 3 ...

    If you aren't sure how to do this, there are 4 possible results (order doesn't matter): p = positive, n = negative

    ppp
    ppn
    pnn
    nnn

    a positive times a positive times a positive is ? and so on:

    ppp = p
    ppn = n
    pnn = p
    nnn = n

    So:

    ppp = pnn
    ppn = nnn

    Remember order doesn't matter. In the example where -1, -2, 3 (nnp) should have been 1, 2, 3 (ppp), did it result in error? No, because ppp = pnn.

    Try the other combinations. Say the correct numbers are ppp.
    What results if exactly 0 are incorrect? ppp. ppp and ppp are equal, so no error.
    If 1 is incorrect? ppn. ppp and ppn are not equal, so this results in error.
    If 2 are incorrect? pnn. ppp and pnn are equal, so no error.
    If 3 are incorrect? nnn. ppp and nnn are not equal, so this results in error.
    Just go in order and do this for the rest, noting when error occurs.
     
    Last edited: May 27, 2005
  11. May 27, 2005 #10
    I see what you guys mean, so am I thinking right?

    For the case, two are correct and one is the wrong number:

    From the 10 numbers, 5 are right and 5 are wrong. Choose one wrong from the 5. so the combination of 5 choose 1 is just 5. Choose 2 right from the other 5 numbers, so the combination of 5 choose 2 is just 10. So there are a total of 5 + 10 = 15 ways for that scenario correct?
     
  12. May 27, 2005 #11
    You are thinking right but should u add 5+10?? or should it be 5*10??
    Once u are done with this, how many in this scenario give u wrong final result??

    -- AI
     
  13. May 27, 2005 #12
    Yeah, it's suppose to be multiplication. My mistake.

    I don't understand why we need to take into account choosing 3 values where 1 or 2 of them willl be incorrect. Because the question just asks how many different ways are there of choosing three numbers at that stage - the stage where no error occurs, which means that 5 numbers that have errors will not be chosen. Can you clear this up for me? I am a bit confused.
     
  14. May 27, 2005 #13
    Nevermind, I mis-read the question. It says the comp chooses 10, so we must take into consideration the comp choosing incorrectly inputted numbers correct?

    Just saw your edited post, honestrosewater
     
  15. May 27, 2005 #14
    right!

    -- AI
     
  16. May 27, 2005 #15
    Ok, after thinking this over, is the following correct?

    Case 1 : all three are correct numbers

    5 choose 3 = 10 ways choosing that will yield no error.

    Case 2 : two are correct and one is the wrong number

    (5 choose 1) * (5 choose 2) = 50 - 1 = 49 ways of choosing that will yield an error.

    Case 3 : one correct and two wrong

    (5 choose 1) * (5 choose 2) = 50 ways of choosing that will not yield an error, since the similar results are achieved.

    Case 4 : all three are wrong

    5 choose 3 = 10 ways of choosing that will yield error.


    In total, there are 61/120 possible ways of choosing. Is this right???????
     
    Last edited: May 28, 2005
  17. May 28, 2005 #16

    honestrosewater

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    Eh, I was hoping someone else would come along, because I don't know that much about probability. I think I do know one way to solve this, but I'm not promising anything.
    You have 10 numbers, 5 correct, 5 incorrect. You will choose 3 numbers without replacement (clue: dependent events). You've figured out which combinations result in error:
    3 correct - no error
    2 correct, 1 incorrect - error
    1 correct, 2 incorrect - no error
    3 incorrect - error.
    You want to know the probability of choosing a combination that doesn't result in error. So what is the probability of choosing A) 3 correct numbers or (clue: addition rule) B) 1 correct and 2 incorrect numbers?

    A) What is the probability of choosing 3 correct numbers?
    You start with 5/10 correct and 5/10 incorrect.
    What is the probability that the first number you choose is correct? 5/10.
    Now 4/9 are correct and 5/9 are incorrect.
    What is the probability that the second number you choose is correct? 4/9.
    Now 3/8 are correct and 5/8 are incorrect.
    What is the probability that the third number you choose is correct? 3/8.
    So the probability of choosing 3 correct numbers is 5/10 * 4/9 * 3/8.

    B) What is the probabilty of choosing 1 correct and 2 incorrect numbers?
    You start with 5/10 correct and 5/10 incorrect...

    Find P(B) and use the addition rule: P(A or B) = P(A) + P(B) - P(A and B).
    Does that make sense?
     
  18. May 28, 2005 #17
    Yes, that makes sense. Thank you. However, I would like to know why we need to get P(B)? Wouldn't the probability of the case for 1 correct and 2 incorrect numbers be 5/10 x 4/9 x 5/8? Or you cannot do this? Furthermore, since these are the two valid cases that will not result in an error, then the total number of ways to choose 3 numbers that will not result in error will be 60/120. Is it possible that it's 50%?


    Again, thanks for your help and patience.


    btw, the probability of P(A intersection B) is just the points that are both common in A and B correct? I have determined the cases where both have common points are ppp and ppn and not pnn and nnn, since there are no two correctly inputted negative numbers. So would P(A intersection B) be just 6000/518400?
     
    Last edited: May 28, 2005
  19. May 28, 2005 #18

    honestrosewater

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    Because you need to find P(A or B).
    You know right away the denominators will be 10, 9, and 8 because you start with 10 numbers and remove 1 number each time (the demoninators represent your sample space).
    I get:
    P(A) = 5/10 * 4/9 * 3/8 = 1/12
    P(B) = 5/10 * 5/9 * 4/8 = 5/36
    I would think A and B are mutually exclusive (you cannot choose both 3 correct and 1 correct, 2 incorrect), so
    P(A or B) = 1/12 + 5/36 = 2/9
    Again, I don't know if this is correct, but here is something to consider:
    There is one way to choose (A) 3 correct and P(A) = 1/12.
    There are 3 ways to choose (C) 2 correct, 1 incorrect and P(C) = 5/36.
    There are 3 ways to choose (B) 1 correct, 2 incorrect and P(B) = 5/36.
    There is one way to choose (D) 3 incorrect and P(D) = 1/12.
    Now adding everything up: 1*(1/12) + 3*(5/36) + 3*(5/36) + 1*(1/12) = 1. This may be what's called the something distribution, and I know somehow something should add to 1, but again, I just don't really know what I'm doing. :redface:
    Someone else may come along soon. Otherwise, I would go with what makes the most sense to you based on what you know. Hope I haven't made things worse. :)
     
  20. May 28, 2005 #19
    Thanks for the help honestrosewater. I understand what you are saying. I probably have to look at this question again and come up with something meaningful.
     
  21. May 28, 2005 #20

    honestrosewater

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    Another way is to just consider that there is/are:
    1 way to choose (A) 3 correct
    3 ways to choose (B) 2 correct, 1 incorrect
    3 ways to choose (C) 1 correct, 2 incorrect
    1 way to choose (D) 3 incorrect.
    So there are 8 possible outcomes and
    P(A) = 1/8
    P(B) = 3/8
    P(C) = 3/8
    P(D) = 1/8
    So P(A or C) = 1/8 + 3/8 = 1/2.
    Maybe the first approach I took was wrong. This one makes more sense. But I'll shut up now. :)
     
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