When must a system of linear equations be solved algebraically.
1 answer:
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Complete Question:
a) List the values that y may take on and construct the pdf table
b) Calculate the average expected matches over the long run of playing this game for the player
c) Calculate the average expected earnings over the long run of playing this game for the player
Answer:
a) Y = { -1, 1, 2,3}
E(X) = 0.5
c) There is a loss of $0.0802
Explanation:
a) Y will take on the values, Y = { -1, 1, 2,3}
The pdf table is attached as a file to this solution
b) E(X) = 0.5
c)
The average expected matches, E(X) = np
where p = 1/6, n =3
E(X) = 3 * 1/6
E(X) = 0.5
c) Average expected earnings =
P(y) can be seen from the pdf table in the file attached
Average expected earning = (-1*0.579) + (1*0.347) + (2*0.069) + (3 * 0.0046)
Average expected earning = -0.0802
This shows that instead of earning, there is a loss of $0.0802
