?Can you solve this probability problem for me please? In the game of roulette,

Question

?Can you solve this probability problem for me please?

In the game of roulette, a player can place a $4 bet on the number 20 and have a 1/38 probability of winning. If the metal ball lands on 20, the player gets to keep the 38 $4 paid to play the game and the player is awarded an additional $140. Otherwise, the player is awarded nothing and the casino takes the player's $4. 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? The expected value is $. (Round to the nearest cent as needed.)

Explanation / Answer

Expected value = 140 * 1/38 + (-4) * 37/38 = $ -0.2105263   (ans)

If game played is 1000, then expected value = -0.2105263 * 1000 = -210.53

So , $210.53 is expected to loose if 1000 games is played

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