In the game of roulette, a player can place a $9 bet on the number 11, and have a 1/38 probability of winning. If the metal ball lands on 11 the player gets to keep the $9 paid to play the game and the player is awarded an additional $315. Otherwise, the player is awarded nothing and the casino takes the player's $9 What is the expected value of the game to the player? If you played the game 1000 times, how much would you expect to lose? What is The expected value: $ (Round to the nearest cent as needed.)

1. What is the expected value of the game to the player?

Expected value= -$0.47368

2. If you played the game 1000 times, how much would you expect to lose?

Expected value= -$473.68

Step-by-step explanation:

1. What is the expected value of the game to the player?

Expected value can tell how much you will gain/lose every time you do the game. To find the expected value you need to multiply every value with its chance to occur. There are two types of events here, winning and losing. You have 1/38 chance of winning $315 and 37/38 chance of losing $9.  

The expected value will be:

expected value= 1/38 * $315 + 37/38*(-$9) = $8.2894 - $8.7631

expected value= -$0.47368

You are expected to lose $0.47368 every time you play

2. If you played the game 1000 times, how much would you expect to lose?

Expected value calculated for each roll. If you want to know how much you gain/lose by multiple rolls, you just need to multiply the expected value with the number of rolls. The game has negative expected value so you are expected to lose money instead of gaining money. The money you expected to lose will be:

number of roll * expected value

1000 timer * -$0.47368/time= -$473.68

You are expected to lose $473.68 if you play the game 1000 times.

We can solve these equations by using substitution method where you substitute one equation to the other and solve for the value of the other variable. We do as follows:

y = 2x^2
y = –3x −1

 2x^2 = –3x −1
 2x^2 +3x +1 = 0

x1= -0.5
x2 = -1
y1 = 0.5
y2 = 2