Answer:
The player's expectation is a loss of 20 cents.
Step-by-step explanation:
In a game of roulette, there are 38 slots where the ball can stop.
The slots are numbered as follows: {0, 00, 1, 2, 3 ..., 36}.
So, N = 38.
The slots which are multiples of 3 are:
S = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33 and 36}
So, n (S) = 12.
Let X = the ball stops on a number that is a multiple of 3
The probability that the ball stops on a number that is a multiple of 3 is:
[tex]P(X)=\frac{n (S)}{N}=\frac{12}{38}=0.32[/tex]
It s provided that the player wins double the amount bet if the ball stops on a number that is a multiple of 3.
Compute the player's expectation as follows:
[tex]E(X)=\sum X\cdot P(X)[/tex]
[tex]=(\$10\times 0.32)+(-\$5\times (1-0.32))\\\\=\$3.20-\$3.40\\\\=-\$0.20[/tex]
Thus, the player's expectation is a loss of 20 cents.
