Question 1194820
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Possible Outcomes:
A = Rolling an even number
B = Rolling a 1, 5, 7 or 13
C = Rolling anything else not mentioned by the previous cases


Here are the various event spaces
A = {2,4,6,8,10,12,14,16,18,20}
B = {1,5,7,13}
C = {3,9,11,15,17,19}


And here are the corresponding probabilities
P(A) = 10/20 = 1/2
P(B) = 4/20 = 1/5
P(C) = 6/20 = 3/10


Thing to notice: P(A)+P(B)+P(C) = 1
The probabilities must sum to 1 in order to have a valid probability distribution. 
Also, the probabilities must be in the interval {{{0 <= p <= 1}}}


Let X = net winnings
This will be the result of subtracting the prize money and the cost to play the game. 


For event A, we have X = 5-3.5 = 1.50
For event B, we have X = 10-3.5 = 6.50
For event C, we have X = 0-3.5 = -3.50


Let's set up a table like so<table border = "1" cellpadding = "5"><tr><td>Event</td><td>Net Winnings</td><td>Probability</td></tr><tr><td>A</td><td>$1.50</td><td>1/2</td></tr><tr><td>B</td><td>$6.50</td><td>1/5</td></tr><tr><td>C</td><td>-$3.50</td><td>3/10</td></tr></table>


Now add on a third column where we multiply the net winnings with their corresponding probability
For example: 1.5*1/2 = 0.75 in the first row.<table border = "1" cellpadding = "5"><tr><td>Event</td><td>Net Winnings</td><td>Probability</td><td>Winnings*Probability</td></tr><tr><td>A</td><td>$1.50</td><td>1/2</td><td>0.75</td></tr><tr><td>B</td><td>$6.50</td><td>1/5</td><td>1.30</td></tr><tr><td>C</td><td>-$3.50</td><td>3/10</td><td>-1.05</td></tr></table>


Adding the results of the third column will get us to the final answer
0.75 + 1.30 + (-1.05) = 1.00


Answer: <font color=red>$1.00</font> is the expected winnings.
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