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Probability mass function for a coin toss

  1. Mar 6, 2012 #1
    1. The problem statement, all variables and given/known data
    A biased coin is tossed ten times. Suppose that the probability of getting heads on a single toss is p. Let X be the number of heads obtained.
    a)Give an algebraic formula for the probability mass function of X.
    b) What do you think E[X] should be.

    2. Relevant equations



    3. The attempt at a solution

    No idea. The probability mass function for a discrete random variable is:

    pX(x)=Pr(X=x) for all x

    I suppose that's what they want for a), but one for this example ? No idea about b), either. It's a biased coin toss. How am I supposed to find this out here? Otherwise, it would be E[X]=0.5
     
  2. jcsd
  3. Mar 6, 2012 #2
    Since they don't tell what you what the bias is, you can kind of ignore it. All you know is that the probability is p. What if you tossed a fair coin 10 times? What is the probability of getting x heads?
     
  4. Mar 6, 2012 #3

    Ray Vickson

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    I do not believe that you are given a problem like this with absolutely no background. Maybe it was not covered in detail in class, but I would bet it is in your textbook, somewhere. If it is not in your textbook and not in your course notes, it is certainly widely available in other books or on line. I think you should make the effort to find it before asking us to do it for you.

    RGV
     
  5. Mar 6, 2012 #4
    As Ray pointed out, this problem is a very basic one.
    However, some points for you (hope you know little probability):
    a)
    1) think of what is the probability that head is obtained once;
    2) then, what is the probability that it is obtained X-times in 10 tosses? You will obtain pmf.

    b) read carefully, how the mean value is defined for discrete distributions. It is very straightforward even to guess the result.
     
  6. Mar 6, 2012 #5
    You are referring to be? I already answered that. It's 0.5. There are only 2 possible outcomes. So the probability is 0.5. Each time there's a 0.5 probability. That's all I can think of.


    It's obviously not "very basic" to me, since I would not have asked then. I don't know what you mean. This dosen't lead me to the answer. My best guess is pX(10)=0.5
    I still don't know how to do b. E[X] is the expected value. 5 heads and 5 tails.
     
  7. Mar 6, 2012 #6
    The probability of getting one head is p, right? What's the probability of getting two heads? Write out the different possibilities, of all the coin tosses. i.e. you can get two heads, a head or tail, a tail or head, or two heads. What are the associated probabilities for each of these conditions?
     
  8. Mar 6, 2012 #7

    Ray Vickson

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    NO, NO, NO! You were *told* the probability of heads is p per toss. Do you still say that the probability of heads in one toss is 0.5? You were told it is not! Suppose I have a really biased coin with probability 0.999 of heads in each toss (and probability 0.001 of tails). Of course, in each trial there are only two outcomes---heads or tails---but that does not mean they are equally likely. They are not: P{heads} = 0.999 and P{tails} = 0.001.

    I cannot figure out whether or not you are reading your textbook or going to class and taking notes. Something is not getting through to you. You are missing the very most basic ideas, and I have no idea how to help you.

    RGV
     
  9. Mar 7, 2012 #8
    XodoX, I did not intend to offend you. Please read twice what the problem is. You have (possibly biased) coin, i.e., Pr(one head in one toss) = p. What is then the probability of one tail in one toss?

    Now think. You have 10 tosses and say you are calculating probability that you obtain right ONE head. This means 1 head AND 9 tails. Answer the following questions:
    1) What is the probability that first 1 hat and then 9 tails occur?
    2) How many events 1H + 9T may happen? Remind, you can get that head in ANY toss, so you will use basic combinatorics.

    Then what rule applies to 2H + 8T. Very probably, you will get an idea what the final answer is.
     
  10. Mar 7, 2012 #9
    It's only a coin with 2 sides. Head and Tail have to have the same probability if I ignore it's biased. Each has a probability of 1/2. Two whatever is then 0.25, and one T or H is 0.5.
     
  11. Mar 7, 2012 #10

    Ray Vickson

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    You are not allowed to ignore the bias: you were told explicitly that the coin is biased. Go back to the example I gave you, where P{heads} = 0.999 and P{tails} = 0.001. In two tosses of this coin, what is the probability of two heads? of two tails? of one head and one tail?

    You seem to be hung up on the false notion that because there are only two outcomes, they must be equally likely. NOT TRUE!

    In fact, even for real coins in the real world there is evidence of bias. See, eg, the articles
    http://www.codingthewheel.com/archives/the-coin-flip-a-fundamentally-unfair-proposition
    and
    http://www.statistik.lmu.de/~helmut/Texte/Euro.pdf [Broken] .
    As this last article points, some of the Euro coins in different countries have different amounts of bias; it also cites experimental work by Diaconis and others, pointing out the existence of bias in some actual coin-tossing situations.

    RGV
     
    Last edited by a moderator: May 5, 2017
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