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Thread: 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 $\displaystyle Y_1$ , the number of heads. Let $\displaystyle Y_2$ 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 $\displaystyle Y_1 \ \ \mbox{and} \ \ Y_2$.

    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.
    $\displaystyle \left[ {\begin{array}{*{20}c} H & H & H \\ H & H & T \\ H & T & H \\
    H & T & T \\ T & H & H \\ T & H & T \\ T & T & H \\ T & T & T \\ \end{array} } \right]$.
    Now $\displaystyle P\left( {Y_1 = 0,Y_2 = - 1} \right) = \frac{1}{8}$ that is the last row in the table.
    Moreover, $\displaystyle P\left( {Y_1 \ne 0,Y_2 = - 1} \right) = 0$. Do you see why?
    $\displaystyle P\left( {Y_1 = 2,Y_2 = 1} \right) = \frac{2}{8}$. That happens in the second and third rows.
    Whereas $\displaystyle P\left( {Y_1 = 1,Y_2 = 2} \right) = \frac{1}{8}$. 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:

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