Answer:
The expected winnings of the player in one game = 2.55
Step-by-step explanation:
Given - A blue die and a red die are thrown in a game. If the sum of the two numbers is 7 or 11, the player wins $10. If the sum of the two numbers is 12, then the player wins $20. In all other cases, the player loses a dollar.
To find - What are the expected winnings of the player in one game ?
Solution -
Given that,
A blue die and a red die are thrown in a game.
The Sample Space becomes -
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (4, 4), (6, 5), (6, 6) }
So,
n(S) = 36
Now,
Let A be the outcome that, the sum of the two numbers is 7 or 11
A = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)
(6, 5), (5, 6) }
So,
n(A) = 8
So,
P(A) = n(A)/ n(S)
= 8/ 36
= 2/9
⇒P(A) = 2/9
Now,
Let B be the outcome that, the sum of the two numbers is 12
B = {(6, 6)}
So,
n(B) = 1
So,
P(B) = n(B)/ n(S)
= 1/ 36
⇒P(B) = 1/36
Now,
Expected value, E(x) = ∑ x P(x)
= 10 × P(A) + 12 × P(B)
= 10 × 2/9 + 12 × 1/36
= 20/9 + 12/36
= (20×4 + 12)/36
= (80 + 12)/36
= 92/36
= 2.55
⇒Expected value, E(x) = 2.55
∴ we get
The expected winnings of the player in one game = 2.55
Simplify