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Probablity Problem 1.3 From Statistical and Thermal Physics, Reif

  1. Aug 23, 2012 #1
    1. The problem statement, all variables and given/known data
    A number is chosen at random between 0 and 1. What is the probability that exactly 5 of its first 10 decimal places consists of digits less than 5?

    2. Relevant equations
    Binomial Coefficient = [itex]\displaystyle \binom{N}{n1}[/itex]
    Where n1 denotes "n1" objects of an indistinguishable type.

    3. The attempt at a solution
    Obviously, 5 digits of the first 10 decimal places of this "number" is less than 5 and the rest are of course greater than 5. Therefore the probability should be [itex]\frac{10!}{10^{10}}.
    The authors solution assumes that there are distinct combinations, and the binomial coefficient (combinations) must be used. In fact his answer is [itex]\displaystyle \binom{10}{5}(1/2)^5(1/2)^5[/itex].

    I do not agree that there are (10 choose 5) choices for the total number of distinct choices. What is the author's question really asking???
  2. jcsd
  3. Aug 23, 2012 #2


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    The number 10C5 refers to the 5 places where the numbers less than 5 will be put. So these numbers can be in places {1,2,3,4,5}, or in {3,4,6,7,8} , etc.

    After the place assignment has been made, there are 5^5 choices for each place. Then the remaining numbers in {5,6,7,8,9} can be made in 5^5 ways, but the

    placements have been fixed.
    Last edited: Aug 23, 2012
  4. Aug 23, 2012 #3

    Ray Vickson

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    Look at a smaller problem: 4 numbers, with two < 5 and two >= 5. Using "L" for "< 5" and "G" for ">= 5", the possible patterns are LLGG, LGLG, LGGL, GGLL, GLGL, GLLG, each having probability (1/2)^2*(1/2)^2 = 1/2^4. There are 6 different patterns, and note that 6 = (4 choose 2).

    In general, for a string of n symbols, k 'L's and (n-k) 'G's, the binomial coefficient (n choose k) counts the number of different strings. You can see this using the argument posted above by Bacle2, or you can verify it directly: let F(n,k) denote the number of distinct strings of length n, with k 'L's and (n-k) 'G's. Clearly, F(n,0) = 1 (there is just one string of all G) and F(n,n) = 1. Also: F(n,1) = F(n,n-1) = n, because if there is just one L (or just one G) it can go in any of the n places. For 2 <= k <= n-1: (i) if L is first, the number of strings is F(n-1,k-1), because that is the number of ways to fill in the rest of the string; and (ii) if G is first, there are F(n-1,k) remaining strings. Altogether, we get F(n,k) = F(n-1,k) + F(n-1,k-1). This is Pascal's triangle, so we get (by induction, for example) that F(n,k) = (n choose k).

  5. Aug 23, 2012 #4


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    Just curious...how did you come up with your answer?
  6. Aug 23, 2012 #5
    I took the total number of arrangements of the first ten digits- ten factorial, then I divided by the total number of possible numbers between 0 and 1 (each digit can be any of the 10 numbers). I am currently studying Bacle2 and Ray's comments on combinations. I am also trying to find the major flaw in my argument.

    Thank you all!
  7. Aug 23, 2012 #6


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    You might want to think about the fact that you didn't use the information that five of the digits were less than 5. How would your answer have changed if the problem had specified four of the digits were less than 5? Or what about if it has asked about five of the digits being less than 1?
  8. Aug 23, 2012 #7


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    I have the impression that when you write 10!, you may be assuming that each digit

    occurs exactly once? That you have 10 digits and you just prove all combinations of

  9. Aug 23, 2012 #8
    The digits don't need to occur once. I forgot the fact that digits are allowed to repeat themselves. I found the flaw.

    Let me try Vela's first question.
    If four of the digits were less than five- than I would multiply the probabilities of four digits less than five (which is (1/2)^4) and rest greater than five ((1/2)^6), and finally multiply by all distinct strings of both types that is (10 choose 4).

    Answer would look like this: [itex]\displaystyle \binom{10}{4}(1/2)^4(1/2)^6[/itex]

    I think I have grasped this problem. Time to re-read everyone's comments and understand a little more about combinatorics. This will be my first time learning about it.
  10. Sep 21, 2012 #9
    For anyone else who might be reading this, I started reading Statistical Physics by Reif (Berkeley Vol. 5) and the same problem appears in the second chapter. Now he's talking in the book about some particles moving in a box among other things and related to the problem I found this type of reasoning pretty good:

    - you can think of the digits as being the particles in a box (10 particles in this case)
    - they can take a spot on the x axis (discrete values here: 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9)
    - if you split the box in two halves you have the particles that can take values: 0, 1, 2, 3, 4 in the left and the ones that take the values 5, 6, 7, 8, 9 in the right

    Then there is the question what is the probability of finding 5 (n) of these 10 (N) particles in the left side of the box (that is on the positions 0, 1, 2, 3, 4) and 5 (N - n) of these particles in the right side of the box (that is on the positions 5, 6, 7, 8, 9).

    Now if you look at only one particle, you have the probability p of it being in the left side and q = 1 - p of it being in the right side. So if you have the box divided in two halves, then p = q = 1/2 because there is no preferential part of the box where the particles would like to be.

    So then, you just have to compute the probability (which is just the definition):

    P(n) = CNn * pn * qN-n = [itex]\frac{10!}{5!(10-5)!}(\frac{1}{2})^{5}(\frac{1}{2})^{5}[/itex]

    I found this to be easier because it uses the same reasoning as in the text (which is about particles confined in a box) when deriving other definitions / equations and it gave me the background to do it. Hope this helps you.
    Last edited: Sep 21, 2012
  11. Jan 19, 2014 #10
    This is really helpful!
    Question: When I first read the problem statement, I interpreted it as being focused on the limiting of the first five positions to numbers in {0,1,2,3,4). Where I went wrong, though, is with the second set of five positions. I missed that there was an implication that the possible numbers would be in {5,6,7,8,9}.

    Apparently I need a more rigorous course in Probability theory...
  12. Oct 14, 2015 #11
    I have a problem with the answers given
    1) clearly in the question is given that "first" five places are filled with the numbers less than five then why 10C5.If we need to fill numbers 0,1,2,3,4 in the places {1,2,3,4,5} than there are 5^5 ways of doing that.
    2) nowhere in question does it ask that the remaining places {6,7,8,9,10} should be filled with numbers greater than 5, they can be filled with numbers less than 5 also so total ways of doing that is 10^5.This makes my answer different from that in the book.plz help.
    Last edited: Oct 14, 2015
  13. Oct 15, 2015 #12
  14. Oct 16, 2015 #13

    Ray Vickson

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    No, you have mis-read the question. It says "exactly five of the first ten digits are less than 5". That means that the other five digits are >= 5; and furthermore, the five digits < 5 can be located anywhere in the first ten spaces.
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