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Math Help - multivariable coin toss question

  1. #1
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    multivariable coin toss question

    Three balanced coins are tossed independently. One of the variable of interest is Y_1 , the number of heads. Let Y_2 denote the amount of money won on a side bet in the following manner. If the first head occurs on the the first toss, you win $1. If the first head occurs on toss 2 or on toss 3 you win $2 or $3, respectively. if no heads appear, you lose $1.

    a) Find the joint probability function for  Y_1 \ \ \mbox{and} \ \ Y_2.

    b) What is the probability that fewer than three heads occur and you will win $1 or less?

    I have no clue on how to do this problem...
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  2. #2
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    First make a table of outcomes.
    \left[ {\begin{array}{*{20}c}   H & H & H  \\   H & H & T  \\   H & T & H  \\<br />
   H & T & T  \\   T & H & H  \\   T & H & T  \\   T & T & H  \\   T & T & T  \\ \end{array} } \right].
    Now P\left( {Y_1  = 0,Y_2  =  - 1} \right) = \frac{1}{8} that is the last row in the table.
    Moreover, P\left( {Y_1  \ne 0,Y_2  =  - 1} \right) = 0. Do you see why?
    P\left( {Y_1  = 2,Y_2  = 1} \right) = \frac{2}{8}. That happens in the second and third rows.
    Whereas P\left( {Y_1  = 1,Y_2  = 2} \right) = \frac{1}{8}. That happens only in the sixth row.

    Now you finish off the table.
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  3. #3
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    if I wanted to derive the marginal probability for the winnings, would it be:

    \left[ \begin{array}{cccc}<br />
-1 & 1 & 2 & 3 \\ \frac{1}{8} & \frac{1}{2} & \frac{1}{4}  & \frac{1}{8} \end{array} \right] ?
    Last edited by lllll; April 7th 2008 at 08:43 PM.
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