In the game of roulette, a player can place a $9 bet on the number 26 and have a StartFraction 1 Over 38 EndFraction probability of winning. If the metal ball lands on 26, 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? Note that the expected value is the amount, on average, one would expect to gain or lose each game.

(a)-$0.47

(b)$4700

Step-by-step explanation:

(a)Therefore:

The expected value of the Game to the Player

The expected value of the game to the player is -$0.47. This means the player is expected to lose $0.47.

(b)If the game is played 1000 times

Expected Loss = 1000 X $0.47

=$4700

Therefore, If you played the game 1000 times, you would be expected to lose approximately $4700.

2? if not the correct answer you might need to provide more information

step-by-step explanation:

step-by-step explanation:

because 1/3 is a fraction. we can tell because there is a divisor symbol between the 1 and the 3. if we were to add 2 fractions together, we would add the numerators, which is the number on top, but not the denominators, which is the number on the bottom

1/3+1/3=2/3

hope this !