Question 1190123
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Part (a)
Event space = {set of 12 face cards} 
Sample space = {set of 52 playing cards}
P(face card) = 12/52 = 3/13
P(non face card) = 1 - P(face card) = 1 - 3/13 = 10/13


Table:<table border = "1" cellpadding = "5"><tr><td>Event</td><td>X</td><td>P(X)</td><td>X*P(X)</td></tr><tr><td>Selects face card</td><td>5</td><td>3/13</td><td>15/13</td></tr><tr><td>Selects something else</td><td>-2</td><td>10/13</td><td>-20/13</td></tr></table>X represents the net winnings for Kyd based on each event.
Negative X values indicate Kyd loses money to North.


Once we computed the X*P(X) column, we add up those items
(15/13) + (-20/13) = (15-20)/13 = -5/13
-5/13 = -0.3846
This rounds to -0.38


Kyd expects, on average, to lose about $0.38 (aka 38 cents) each time he plays the game.


Answer:  <font color=red>-0.38 dollars</font>


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Part (b) 


If Kyd expects to lose 38 cents on average, per game, then North expects to win 38 cents on average. 


You could go through the same steps as part (a), but flip each item in the X column. That will flip each item in the X*P(X) column as well.
Adding said items in the X*P(X) column should get you 5/13 = 0.38 approximately.


Answer:  <font color=red>0.38 dollars</font>
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