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rise, the player is awarded nothing and the casino takes the player's $6. Find the expected value E(x) to the player for one play of the game. If x is the gain to a player in a game of chance, then E(x) is usually negative. This value gives e amount per game the player can expect to lose. pected value is $. d to the nearest cent as needed.)

rise, the player is awarded nothing and the casino takes the player's $6. Find the expected value E(x) to the player for one play of the game. If x is the gain to a player in a game of chance, then E(x) is usually negative. This value gives e amount per game the player can expect to lose. pected value is $. d to the nearest cent as needed.)

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.8: Probability
Problem 68E
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In the game of roulette, a player can place a $6 bet on the number 17 and have a 1 probability of winning. If the metal ball lands on 17, the player gets to keep the $6 paid to play the game and the player is awarded an additional $210. 38 Otherwise, the player is awarded nothing and the casino takes the player's $6. Find the expected value E(x) to the player for one play of the game. If x is the gain to a player in a game of chance, then E(x) is usually negative. This value gives the average amount per game the player can expect to lose. ... The expected value is $ (Round to the nearest cent as needed.)
Transcribed Image Text:In the game of roulette, a player can place a $6 bet on the number 17 and have a 1 probability of winning. If the metal ball lands on 17, the player gets to keep the $6 paid to play the game and the player is awarded an additional $210. 38 Otherwise, the player is awarded nothing and the casino takes the player's $6. Find the expected value E(x) to the player for one play of the game. If x is the gain to a player in a game of chance, then E(x) is usually negative. This value gives the average amount per game the player can expect to lose. ... The expected value is $ (Round to the nearest cent as needed.)
Expert Solution
Step 1

Given

Probability of winning a game = 1/38

Here a person bet on a particular number 17 of rupees $6

So when a metal ball lands correctly on 17 then he need to pay $6 to play the game

 

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