Answer:
Expected value of profit is ≅ $15.80
Step-by-step explanation:
from the Question,
Let the variable X represents the expected value of profit. X is called as random variable because picking a number from 000-999 digits is a Random process.
P(win) = [tex]0.001[/tex]
So, P(lose) = [tex]1-0.001=0.999[/tex]
Suppose that,
we really want to win this lottery. so we can go to the store and spend $1000 to buy all ticket (from 000 - 999). This would ensure your winning of $500 with one of the tickets (for a $499 profit), but the other 999 would be losers (for a $999 loss).
What would be your average winnings on a per-ticket basis?
[tex]u = E(x) =[/tex]∑[tex]x\times P(x)[/tex]
[tex]= 499\times0.0001+ (-1)\times0.999[/tex]
= [tex]-0.50[/tex]
Here,
Standard deviation of the expected winnings
[tex]V (X) =[/tex] ∑[tex](x-u)^{2} \times P(x)[/tex]
= ∑ [tex](499-(-0.50))^{2} \times 0.001 + (-1-(-0.50))^{2} \times 0.999[/tex]
[tex]= 249.75[/tex]
Taking square root of the variance to get the standard deviation:
SD(x) = [tex]\sqrt{249.75}[/tex]
≅ $15.80
Hence
The expected value of profit is ≅ $15.80