Question

Suppose you open a new game at the county fair. When patrons win, you pay them $3.00; when patrons lose, they pay you $1.00. If the probability of a patron winning is p = .20, then how much can you expect to win (or lose) in the long run? Hint: You need to compute the expected value of the mean.A) win 0.20 cents per playB) win 0.60 cents per playC) lose 0.80 cents per playD) lose 2.20 dollars per play

Answer 1

Answer:

$E(X) = 1 *0.8 - 3*0.2 = 0.2$

so at the long run we can conclude that the best option is :

A) win 0.20 cents per play

Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).  

Solution to the problem

Let X the random variable who represent the ampunt of money win/loss at the game defined.

The probability of loss $3.00 for this game is 0.2 and the probability of win is 1-0.2=0.8 and you will recieve $1.00 if you win. The expected value is given by:

$E(X) = \sum_{i=1}^n X_i P(X_i)$

And for this case if we replace we got:

$E(X) = 1 *0.8 - 3*0.2 = 0.2$

so at the long run we can conclude that the best option is :

A) win 0.20 cents per play