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Probability Problem (unfair coin)

  1. Sep 10, 2006 #1
    Probability (unfair coin)

    Hello everyone, I need a little help with this problem I have.

    A unfair coin is flipped 3 times.
    probability of Head = 2(probability of a tail)

    a. How many different outcomes are there?
    b. How many different outcomes for the # of heads?
    c. Is exactly 2 heads more likely than 3 of a kind?

    Is there a equation to this problem, if someone can assist please help, thanks for looking.
    Last edited: Sep 10, 2006
  2. jcsd
  3. Sep 10, 2006 #2
    a) If you flip a coin once, what are the possible outcomes? Either Heads or Tails right? If you flip a coin twice, the possible outcomes are: (H,H), (H,T), (T,H), (T,T). Where "H" stands for "heads" and "T" stands for "tails" and where (H, H) means that the outcome of the first coin flip was heads, and the second coin flip was also heads. So what now are the possible outcomes of flipping a coin 3 times?

    b) let us go back to the example of flipping a coin twice to give you an idea of what to do when you have to flip the coin three times . If you flip the coin twice, you can have a case where you get no heads (ie, the case (T,T), and thus # of heads = 0), you can also get the case where the # of heads = 1 (ie, the cases, (H,T), and (T,H)), and you can get a case where the # of heads = 2 (ie, the case (H,H)). So the number of heads is going to be either 0, 1, or 2.

    c) let us, again, look at flipping the coin only twice, and then you can figure out what to do when you have to flip the coin three times. In this case let us change the problem to the following: Is getting exactly one head more likely than 2 of a kind?

    First, with your unfair coin, the probability of the coin landing on heads is P(H) = 2*P(T), (that is, 2 times the probability of landing on tails).

    Recall, the probabilities of exhausitve and mutually exclusive events must add to 1. So, P(H) + P(T) = 1, but P(H) = 2*P(T), so P(H) + P(T) = 2*P(T) + P(T) but this = 1. So 3*P(T) = 1, thus P(T) = 1/3, and thus P(H) = 2/3.

    So P(exactly 1H) = P((H,T)) + P((T,H)) = (2/3)*(1/3) + (1/3)*(2/3) = 2/9 + 2/9 = 4/9
    and P(exactly 2H) = P(H,H) = (2/3)*(2/3) = 4/9.

    So in this case, getting exactly one head is not more likely than getting 2 of a kind (here the likelihood of getting both is actually the same).
    Last edited: Sep 10, 2006
  4. Sep 11, 2006 #3
    I just want to point out in terms of what has been correctly written by mattmns, that I find it easier to consider these problems in terms of binominal expansion. Here we have p=2/3, q=1/3.

    In the case of two flips we have (p+q)^2 = p^2+2pq+q^2.

    P^2 represents the result of two heads, equal to (2/3)^2 = 4/9 and the result of both a head and tale is 2pq=2(2/3)(1/3) = 4/9. (Since (p+q)^2 = 1^2 =1, we know that all cases add to 1.)

    The fact that this can be seen as binominal expansion is because in the above pp occurs once, but pq can also occur as qp, so the use of 2, and so forth.
  5. Sep 11, 2006 #4
    Thanks for the post.
    This is for question a, so the probability for different outcomes for an unfair coin would not matter if it was a fair coin, the probability would be the same as a fair coin?

    This is for question b, fair or unfair coin, the # of outcome of heads would not matter? I am just a little comfused.
  6. Sep 11, 2006 #5


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    Note that the answers to a) and b) have nothing to do with the fact that this is an "unfair" coin. On each toss, the coin still can only be Heads or Tails, regardless of probability. There are still 23= 8 possible outcomes (not "equally likely").

    I would rephrase that- as far as the # of outcomes of heads was concerned, "fair" or "unfair" does not matter. As far as the number of heads is concerned there are still 4 possiblities, 0, 1, 2, 3.
    Last edited: Sep 12, 2006
  7. Sep 11, 2006 #6
    Thanks for posting and helping me with my question. I understand now for question a and b, but for c I have to look in to binominal expansion, I have to look into it a little more. Thanks again.
  8. Sep 12, 2006 #7


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    Science Advisor

    Since probability of Head = 2(probability of a tail) but probability of Head + probability of tail= 1 (one one flip one of those must happen), letting p= probability of tail, 2p+ p= 1 so p= 1/3. Probability of a Tail is 1/3, probability of a Head is 2/3.

    The probability of "exactly 2 heads in 3 flips" is

    The probability of "all 3 flips heads" is
    and the probability of "all 3 flips tails" is
    Of course, the probability of "three of a kind" is the sum of those.
  9. Sep 15, 2006 #8
    Thanks everyone for helping me out with this question, now I just have one more question, what is the equation for combination. I have an idea what it is, I just need to confirm it. Thanks for the help.
  10. Sep 15, 2006 #9
    Equation for combination? Do you mean: [tex] \left( \begin{array}{cc}n \\ r \end{array} \right) [/tex] ???

    If that is what you are talking about, then

    [tex] \left( \begin{array}{cc}n \\ r \end{array} \right) = \frac{n!}{r!(n-r)!} [/tex]
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