# Thread: multivariable coin toss question

1. ## 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...

2. First make a table of outcomes.
$\left[ {\begin{array}{*{20}c} H & H & H \\ H & H & T \\ H & T & H \\
H & T & T \\ T & H & H \\ T & H & T \\ T & T & H \\ T & T & T \\ \end{array} } \right]$
.
Now $P\left( {Y_1 = 0,Y_2 = - 1} \right) = \frac{1}{8}$ 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.

3. if I wanted to derive the marginal probability for the winnings, would it be:

$\left[ \begin{array}{cccc}
-1 & 1 & 2 & 3 \\ \frac{1}{8} & \frac{1}{2} & \frac{1}{4} & \frac{1}{8} \end{array} \right]$
?