In the game of​ roulette, a player can place a ​$ bet on the number and have a probability of winning. If the metal ball lands on ​, the player gets to keep the ​$ paid to play the game and the player is awarded an additional ​$. ​ Otherwise, the player is awarded nothing and the casino takes the​ player's ​$. 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?

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The expected amount to be lost is  0.5263 cents per game

b

The expected amount to be lost is  

     [tex]P = \$ 526 .3 [/tex]

Step-by-step explanation:

From the question we are told that

   The probability of winning is  [tex]p = \frac{1}{38} = 0.0263[/tex]

Generally the probability of losing is mathematically evaluated as

            [tex]q =  1-p [/tex]

=>        [tex]q = 1-0.0263 [/tex]

=>        [tex]q = 0.9737[/tex]

Generally the expected value is mathematically represented as

      [tex]E(X) =  \sum x_i *  P(x_i)[/tex]

Here [tex]x_i[/tex] is a discrete variable (i.e it is a countable )  are outcomes of the player winning and the player losing

So

     [tex]E(X) =   x_1 *  P(x_1) + x_2 *  P(x_2)  [/tex]

Now the outcome of winning is  making $350  so

       [tex]x_1 = \$ 350[/tex]

        the outcome of losing is losing $10  so  

        [tex]x_2 = - \$10[/tex]

So

     [tex]E(X) =  350* 0.0263 -10 *  0.9737  [/tex]

       [tex]E(X) = -0.5263 [/tex]

hence the expected amount to be lost is  0.5263 cents per game

If you played the game 1000​ times, the amount that is expected to be lost is  

    [tex]P =  1000 * -0.5263[/tex]

=> [tex]P = -\$ 526 .3 [/tex]

Hence the expected amount to be lost is  

     [tex]P = \$ 526 .3 [/tex]

4 log x = 4

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