Respuesta :
Answer:
Step-by-step explanation:
The main idea is that I would like to pay less than what I'm expecting to win, so in that way, I get a profit out of playing this game. Let X be the number of tosses until I get a Heads. By definition, this is a geometric random variable with parameter p = 1/2.
Let Y the amount I received for playing. So, we want to calculate the expected value of Y.
We can calculate it as follows
[tex]E[Y] = 2 P(X=1)+ 4 P(X=2)+ 8 P(X =3) + \dots = \sum_{n=1}^infty 2^n P(X=n)[/tex]
Since X is a geometric random variable, we have that [tex] P(X=n) = (\frac{1}{2})^{n-1}\frac{1}{2}[/tex]
Then,
[tex] E[Y] = \sum_{n=1}^\infty 2^{n} (\frac{1}{2})^{n-1} \frac{1}{2} = \sum_{n=1}^\infty 2^{n-1} \cdot \frac{1}{2^{n-1}} = \sum_{n=1}^\infty 1 = \infty[/tex]
So, we expect to have an infinite amount. Given this, we can pay as much as we want to play the game.
