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Math Help - probability mass function..

  1. #1
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    probability mass function..

    stuck on this too

    "In the game of Risk, battles are decided by the rolling of dice. Suppose that there are two armies,
    red and blue. The red army rolls three dice and the blue army rolls two dice. Whichever army rolls
    the highest number (on a single die) is declared the winner. In the event of a tie (both armies have
    the same highest number), the blue army is declared the winner.

    (a) Let X denote the highest number rolled by the red army. Find the probability mass function
    of X.
    (b) Let Y denote the highest number rolled by the blue army. Find the probability mass function
    of Y .
    (c) Calculate the probability that the red army wins."


    thanks peeps
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by situation View Post
    stuck on this too

    "In the game of Risk, battles are decided by the rolling of dice. Suppose that there are two armies,
    red and blue. The red army rolls three dice and the blue army rolls two dice. Whichever army rolls
    the highest number (on a single die) is declared the winner. In the event of a tie (both armies have
    the same highest number), the blue army is declared the winner.

    (a) Let X denote the highest number rolled by the red army. Find the probability mass function
    of X.
    (b) Let Y denote the highest number rolled by the blue army. Find the probability mass function
    of Y .
    (c) Calculate the probability that the red army wins."


    thanks peeps
    Well what have you done? Try starting with part (a)

    CB
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    Well what have you done? Try starting with part (a)

    CB

    that is kind of the problem, im not sure exactly how to find the pmf of each of the cases, i.e. not sure where to start to be honest...
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by situation View Post
    that is kind of the problem, im not sure exactly how to find the pmf of each of the cases, i.e. not sure where to start to be honest...
    Number the dice, then the probability that the first has the highest (or equal) score and that is 3 is (1/6)(3/6)(3/6). This is the probability that die 1 shows a 3, times the probability that die 2 shows 1,2 or 3, times the probability that die 3 shows a 1,2 or 3.


    Now to get the probability that the highest score is 3 we multiply this by the number of permutations of three objects: 3!.

    Hence the probability that the highest roll on three dice is 3 is: 3!(1/6)(3/6)(3/6).

    Now that leaves you to do 1,2,4,5,6.

    CB
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