Probability Functions

In this section we provide a brief overview of those concepts in the theory of probability that are useful for our study of statistics, as well as basic concepts of probability distributions, both discrete and continuous.

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6 Responses to Probability Functions

  1. Katya Chipriyanova Todorova says:

    Ich wohne jetzt in Germany. Ich kann nicht schreiben auf englisch, aber ich kann lesen. In Bulgarien war ich Mathelehrerin. Mir ist interessant noch Mathematik und Statistik. macht mich spass lesen. Danke, Herr Zaiontz!
    Katya todorova

  2. Peter Duncan says:

    Dear Charles,

    I apologise for posting this here. I was not sure where my question would fit best.

    I have the following difficult problem that I would like to solve and I hope that you can help me, please.

    I would like to know which formula to use for the following combinations, please.

    A or B can be chosen x times to generate a series of length x. A and B can be chosen as often as you like. Order does not matter. The question for each series is: For what fraction of the total are there more A’s than B’s?

    Examples:
    If x=2, possibilities are A-A, A-B, B-A, B-B so how often are there more A’s than B’s? That is 1 out of 4=0.25.

    If x=4, possibilities are A-A-A-A, A-A-A-B, A-A-B-A, A-A-B-B, A-B-A-A, A-B-A-B, A-B-B-A, A-B-B-B, B-A-A-A, B-A-A-B, B-A-B-A, B-A-B-B, B-B-A-A, B-B-A-B, B-B-B-A, B-B-B-B
    so how often are there more A’s than B’s? That is 5 out of 16=0.3125.

    I can do these by hand, but it becomes more difficult when x=62.

    Note that I am not giving any examples where x is an odd number, because that answer would always be 0.5.

    A formula would be sooooo helpful!

    Thank you,
    Peter

    • Charles says:

      Peter,
      Let pA = the probability of more A’s than B’s. Let pB = the probability that there are more B’s than A’s. Let pAB = the probability that there are the same number of A’s as B’s. Thus pA + pB + pAB = 1 and pA = pB. Thus if you can calculate pAB, you are done. I’ll leave calculating pAB to you, but you might want to look at the binomial distribution.
      Charles

  3. Peter Duncan says:

    Hi Charles,

    Thank you for your suggestion. I found this very complicated, but a person named Henry was able to give me the answer I was looking for. Please see here:
    https://math.stackexchange.com/questions/2600972/problem-with-combinations?answertab=votes#tab-top

    Thank you,
    Peter

    • Charles says:

      Peter,

      This is the solution that I suggested to you, in my response repeated below:

      “Let pA = the probability of more A’s than B’s. Let pB = the probability that there are more B’s than A’s. Let pAB = the probability that there are the same number of A’s as B’s. Thus pA + pB + pAB = 1 and pA = pB. Thus if you can calculate pAB, you are done. I’ll leave calculating pAB to you, but you might want to look at the binomial distribution.”

      Thus, pA + pA + pAB = 1 and so pA = (1-pAB)/2. Using the binomial distribution, pAB = C(2n,n)/2^(2n) where 2n is the number of As and Bs.
      Thus pA = (1-C(2n,n)/2^(2n))/2

      Charles

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