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Homework Help: Discrete Random Variables

  1. Nov 22, 2004 #1

    Mo

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    Hello (first time poster),
    i am having quite a bit of trouble with a particular problem on stats (which i despise of!) - in particular, discrete random variables.Ok here is the question:


    "Find the probability distribution of X in each of the following questions ...

    Two fair dice are thrown. X is the smaller of the two scores on the dice"


    I have managed to do the first few, but this seems to have thrown me a little bit.I would be grateful for any help indeed.If anyone knows of any good resources online for AS-level Stats, i would also be grateful for the link.

    Regards,
    Mo
     
  2. jcsd
  3. Nov 22, 2004 #2
    Is that a question or part of what is given?

    If you want the probability that a particular die is the smaller of two, you would just have to count all the possibilities for the dice not being equal and divide by 36. There are 6 cases that the dice are equal, this leaves 30. You only want half of these because in 15 cases X is smaller, in the other 15, the other die is smaller. So it is 15/36.
     
  4. Nov 22, 2004 #3

    Galileo

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    Hi.

    It's not too hard to find P(X=a) for a certain a. You just count all the outcomes which have a as a lowest score and divide by 36.

    For example: P(X=3)=7/36.
    (7 outcomes: (3,3),(3,4),(3,5),(3,6),(4,3),(5,3),(6,3))
     
  5. Nov 22, 2004 #4

    Mo

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    That is a whole question (well, sub-question) Believe it or not, i think i almost had it. here was my working out for the question:

    X = Smaller of two scores

    SET =
    1,1 1,2 1,3 1,4 1,5 1,6
    2,1 2,2 2,3 2,4 2,5 2,6
    3,1 3,2 3,3 3,4 3,5 3,6
    4,1 4,2 4,3 4,4 4,5 4,6
    5,1 5,2 5,3 5,4 5,5 5,6
    6,1 6,2 6,3 6,4 6,5 6,6

    i then carried on with ..

    P (X=1) P(1, any other except for 1) =5/36
    P (X=2) P(2, any other except for 1 or 2) =4/36
    P (X=3) P(3, any other except for 1,2 or 3) =3/36
    P (X=4) P(4, 5 or 6) =2/36
    P (X=1) P(5, 6) =1/36

    Where i was getting confused was because i had been taught that the probabilities should add always add upto 1?

    Thanks for your help so far :)
     
  6. Nov 22, 2004 #5
    You did have it... you just need to add up your P(x=1)+...+P(x=5) to get 15/36.

    P(X is smaller) + P(X isnt smaller) = 1 = 100%, thats when it adds to 1...
     
  7. Nov 22, 2004 #6

    Mo

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    Ahh, yes, as in the total probabaility will be equal to 1.Thanks for your help, it is much appreciated.
     
  8. Nov 22, 2004 #7
    He needs it for a particular die and equal shouldn't be counted, so P(X=3) = 3/36
     
  9. Nov 22, 2004 #8

    Galileo

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    That's not what the question asks.

    A random variable is a function from the sample space to [0,1].
    If the event is throwing a 3 and a 6 - (3,6), then X=3. If the event is (2,4), then X=2.
    And ofcourse (3,3) gives X=3.
    The probability of getting X=3 is 7/36 because of the reason I previously gave.

    He is asked for the distrubution function, which according to my definition is:
    [tex]F(a)=P(X\leq a)[/tex]
    although I think the term is easily confused with the probability mass function:
    [tex]p(a)=P(X=a)[/tex]
     
  10. Nov 22, 2004 #9
    I didn't realize that probability distribution was a function, guess I shouldnt have answered...

    He thought it should have added to 1, your way does so thats probably why he thought that. Sorry Mo...
     
  11. Nov 24, 2004 #10

    Mo

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    Thanks for your help both.I completed the answer below just for anyone's future reference:


    X = Smaller of two scores

    SET =
    1,1 1,2 1,3 1,4 1,5 1,6
    2,1 2,2 2,3 2,4 2,5 2,6
    3,1 3,2 3,3 3,4 3,5 3,6
    4,1 4,2 4,3 4,4 4,5 4,6
    5,1 5,2 5,3 5,4 5,5 5,6
    6,1 6,2 6,3 6,4 6,5 6,6

    (another, most probabaly easier method of viewing would be ...)

    1 2 3 4 5 6
    1 1 1 1 1 1 1
    2 1 2 2 2 2 2
    3 1 2 3 3 3 3
    4 1 2 3 4 4 4
    5 1 2 3 4 5 5
    6 1 2 3 4 5 6

    (basically this shows which number is the smallest out of the pair (from both die))

    i then carried on with ..

    P (X=1) P(where 1 is smallest number) =11/36
    P (X=2) P(where 2 is smallest number) =9/36
    P (X=3) P(where 3 is smallest number) =7/36
    P (X=4) P(where 4 is smallest number =5/36
    P (X=5) P(where 5 is smallest number) =3/36
    P (X=6) P(where 6 is smallest number) =1/36

    Notice how they do all add up to 1 (36/36)

    Regards,
    Mo
     
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