in a game of roulette a player can place a 4 dollar bet on n

in a game of roulette a player can place a 4 dollar bet on number 15 and have 1/38 probability of winning.if the metal ball lands on 15 the players get to keep 4 dollars paid to play the game and the player is awarded an additional 140 dollars otherwise the player is awarded nothing and the casino takes the players 4 dollars . what is the expectd value of the game to the player? if you played the game 1000 times, how much would you expect to lose

the expected value is $ (round to the nearest cent as needed)

the player would expect to losse about $ (round to the nearest cent as needed)

Solution

The expectd value of the game to the player,

E(X) = 140 * 1/38 + (-4) * 37/38

        = -4/19 = -0.2105 dollar   (ANS)

If you played the game 1000 times, expect to loose = 1000 * 0.2105 = 210.5 dollars   (ANS)