Probability mass function

Discussion in 'Physics & Math' started by kingwinner, Oct 30, 2008.

  1. kingwinner Registered Senior Member

    Messages:
    796
    1) Suppose we roll a fair die. Whatever number comes up, we toss a coin that many times. What is the probability mass function of the number of heads?


    Let X=number of heads
    Then I know that X can take on the values 0,1,2,3,4,5,6.
    How to proceed from here I have no clue...

    Could someone please explain? Any help would be appreciated!
     
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  3. Pete It's not rocket surgery Registered Senior Member

    Messages:
    10,167
    There's probably (P=1) a more elegant way, but I'd do a separate table for each die result, then multiply each item in each table by 1/6 to get a combined table... if that makes sense.
     
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  5. kingwinner Registered Senior Member

    Messages:
    796
    How about the coin???

    By the way, is this a "birvariate" question? How can I define the second random variable?
     
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  7. Pete It's not rocket surgery Registered Senior Member

    Messages:
    10,167
    I meant that for each roll of the die, draw a separate probability table for the coin flips.
    For example:
    Code:
    4 on die:
    Heads |  P
      0   | 0.0625
      1   | 0.25
      2   | 0.375
      3   | 0.25
      4   | 0.0625
    Then combine the tables, dividing each of the listed probabilities by six and adding matching items from each table.
    For example, to get the combined probability for 2 heads, you'd divide the "2 head" probability from each table by six and add them together:
    Code:
    2 on die:
    Heads |  P
      2   | 0.25
    
    3 on die:
    Heads |  P
      2   | 0.375
    
    4 on die:
    Heads |  P
      2   | 0.375
    
    5 on die:
    Heads |  P
      2   | 0.3125
    
    6 on die:
    Heads |  P
      2   | 0.234375
    
    
    Total:
    ---------------
    Heads |  P
      2   | 0.2578125
    (I think that's right... but you'd better check!)

    It does appear to be a bivariate problem, with the two variables being the die roll and the coin flip.
    But I'm not sure of what that means for the "correct" way to approach the problem. I'm approaching in a brute force intuitive kind of way.
     
    Last edited: Oct 31, 2008
  8. paulfr Registered Senior Member

    Messages:
    227
    This is the binomial probability distribution because the coin toss satisfies the conditions for a Bernoulli Trial.
    1/ Binomial outcome ; H or T
    2/ Each toss is independent of the other events

    P(r) = C(n,r) p^n-r (1 - p)^r
    Plot P(r) verses r to get the distribution
    n = the result of the die roll

    If the die comes up 4, n = 4 and r is 0, 1,2,3,4, the possible outcomes for which the probability is determined by the distribution above.
     
  9. Vern Registered Senior Member

    Messages:
    695
    Doesn't the 0 need to become a 1 since there is no 0 face on the die.
     
  10. Steve100 O͓͍̯̬̯̙͈̟̥̳̩͒̆̿ͬ̑̀̓̿͋ͬ ̙̳ͅ ̫̪̳͔O Valued Senior Member

    Messages:
    2,346
    No. That is how many heads are flipped.
     
  11. Vern Registered Senior Member

    Messages:
    695
    Ok; now it makes sense; reading too fast

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