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Coin Flip Probability

  1. Sep 14, 2011 #1
    A biased coin that lands heads with probability 0.6 is flipped 2 times.

    a. What is the probability of getting an even number of heads?
    b. Given that more heads than tails appear, what is the probability that all of the flips are H?
    c. same as (a) except now the coin is flipped 10 times instead of 2
    d. same as (b) except now the coin is flipped 10 times instead of 2


    So for this problem I figure there are still 4 possible outcomes for a and b....being HH HT TH TT.
    For even number of heads, HH or TT must happen.

    So is the probability = P(HH) + P(TT) / Possible outcomes?


    and for B....since more Heads is the condition, then P(HH) = 1.

    C and d I'm not sure how to start since there are 2^10 possible outcomes right?
     
  2. jcsd
  3. Sep 14, 2011 #2

    gb7nash

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    You don't need to divide by anything. Since these are mutually exclusive events, just simply add the probabilities together.

    Correct

    Do you know what a bernoulli trial is? If not...

    http://en.wikipedia.org/wiki/Bernoulli_trial

    Same idea here. Find the probability for each valid number of heads and add them together.
     
  4. Sep 14, 2011 #3

    Ray Vickson

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    You have stated the result for (b) badly: it is not true that P{HH} = 1, but it is true that P{HH|more heads} = 1. It is a _conditional_ probability.

    As for (c) and (d): have you studied the binomial distribution yet?

    RGV
     
  5. Sep 14, 2011 #4
    Ok so let see if I finally get this.

    So for (A) I add up the probability of getting HH and TT
    Which would be (.6)^2 + (0.4)^2

    For (C) now there are 6 possibilities to get an even number of heads.
    So does that mean its simply (10C6) (.6)^6*(.4)^4

    I have a feeling that this is wrong and I'm calculating the probability of getting 6 heads....not sure :(

    and again lost I'm lost for D since not sure about C
     
  6. Sep 14, 2011 #5
    Or I suppose what would make more sense would be:
    (10C0) (.6)^10*(.4)^0 + (10C2) (.6)^8*(.4)^2 +......(10C10) 1*(.4)^10 ?
     
  7. Sep 15, 2011 #6

    PeterO

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    That looks better for (c).

    Using binomial again you have a big fraction.
    Denominator is Sample Space - all those cases where the number of heads exceeds the number of tails.
    Numerator = event space - which is the one(s) which have all heads.
     
  8. Sep 15, 2011 #7
    Are you saying that with 2 flips of the coin and getting no heads ie TT is an even number of heads?

    And with 2 flips would you agree or not agree that cases a) and b) are the same?
     
  9. Sep 15, 2011 #8

    gb7nash

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    Yes, since 0 heads is an even number of heads.

    They are not the same. For a), you're allowing 0 or 2 heads. For b), you're allowing more heads than tails, i.e. 2 heads.
     
  10. Sep 15, 2011 #9
    Well, I would have taken HH as being 2 head flips and TT as being no head flips, in which case a) and b) would be the same.
    I agree, with HH and TT as even heads interpretation they are not the same.
     
  11. Sep 16, 2011 #10

    vela

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    Even with your interpretation, the two probabilities aren't the same. Part (b) is asking for the probability of two heads given that you know that there were more heads than tails. It's a conditional probability.
     
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