Answer:
Step-by-step explanation:
Given that:
To bet $5 that the outcome is any one of these five possibilities: 0, 00, 1, 2, 3.
Let Y represent the Amount of net profit
Then, Y= {-5, 30}
The probability distribution of Y is:
Y -5 30
P(Y=y) [tex]\dfrac{33}{38}[/tex] [tex]\dfrac{5}{38}[/tex]
a) The expected value of X is given by:
[tex]E[Y] =\sum y P(Y=y)= 30*\dfrac{5}{38}-5*\dfrac{33}{38}[/tex]
[tex]=-0.39[/tex]
[tex]= -39 \ cents[/tex]
b)
On a bet of $5 on the number 25 we are expected to loose 24 cents.
While on a $5 bet that the outcome is any one of the numbers 0,00, or 1 we are expected to loose 39 cents.
Hence, $5 bet on the number 27 is better. Because the expected loss is less in this bet