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)

a) The expected value of X is given by:
![E[Y] =\sum y P(Y=y)= 30*\dfrac{5}{38}-5*\dfrac{33}{38}](https://tex.z-dn.net/?f=E%5BY%5D%20%3D%5Csum%20y%20P%28Y%3Dy%29%3D%2030%2A%5Cdfrac%7B5%7D%7B38%7D-5%2A%5Cdfrac%7B33%7D%7B38%7D)


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
Answer:
6.8818249813
Step-by-step explanation:
I didn't really get the problem but basically, according to "pemdas" you have to multiply 7x1910 and that would be 13370 and next you would do 92010 divided by 13370 which is 6.8818249813.
Since it's p over four, which is a fraction, multiply each side by four. p = 20
7x-5=5x+23
7x-5x=23+5
2x=28
x=28/2
x=14