In the game of? roulette, a player can place a $9 bet on the number 19 and have

Question

In the game of? roulette, a player can place a $9 bet on the number 19 and have a 1/38 probability of winning. If the metal ball lands on

19, 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.

The expected value is ?$nothing .

?(Round to the nearest cent as? needed.)

Explanation / Answer

Solution:-

=> E(x) = 315(1/38) + (-9)37/38) = (315 - 333)/38 = -0.4737

=> 1000*-0.4737 = $-473.70

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.