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Harlamova29_29 [7]
2 years ago
10

In the game of​ roulette, a player can place a ​$5 bet on the number 11 and have a startfraction 1 over 38 endfraction probabili

ty of winning. if the metal ball lands on 11​, the player gets to keep the ​$5 paid to play the game and the player is awarded an additional ​$175. ​ otherwise, the player is awarded nothing and the casino takes the​ player's ​$5. 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?
Mathematics
1 answer:
Sauron [17]2 years ago
4 0
We know that
P(winning) = 1/38 
P(losing) = 1-1/38 = 37/38 
Expected value = 175*1/38 - 5*37/38 
175/38 - 185/38 = - 10/38 =-$0.2632

the answer part A) is
The player's expected value is -$0.2632 ( is losing)

b) <span>if you played the game 1000​ times, how much would you expect to​ lose?
</span>-$0.2632*1000=-$263.20

the answer Part B) is $263.20 
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