After paying ​$5 to​ play, a single fair die isa single fair die is ​rolled, and you are paid back the number of dollars corresponding to th

Question

After paying ​$5 to​ play, a single fair die isa single fair die is ​rolled, and you are paid back the number of dollars corresponding to the numberthe number of dots facing up. For​ example, if aa 66 turns​ up, ​$66 is returned to you for a net​ gain, or​ payoff, of​ $1, if aa 44 turns​ up, ​$44 is returned for a net gain of minus−​$1, and so on. What is the expected value of the​ game? Is the game​ fair?

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Raelynn 1 week 2021-11-25T17:59:49+00:00 1 Answer 0 views 0

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    2021-11-25T18:01:22+00:00

    Answer:

    E(X) = -1.5

    No, the game is not fair since the expected value represents a loss.

    Step-by-step explanation:

    The expected value of this game is calculated using

    E(X) = \sum(x_{i}p_{i})

    Where x_{i} is the net gain of each outcome and p_{i} is the probability of each outcome.

    When a die is rolled, the probability of getting each outcome is

    p = \frac{1}{6}

    Where 6 is the total number of possible outcomes.

    The cost of playing the game is $5

    The net gain for each outcome is given by

    x_{1} = 1 - 5 =-4\\x_{2} = 2 - 5 =-3\\x_{3} = 3 - 5 =-2\\x_{4} = 4 - 5 =-1\\x_{5} = 5 - 5 = 0\\x_{6} = 6 - 5 =1\\

    Now we can find the expected value of this game,E(X) = (x_{1} \cdot p_{1}) + (x_{2} \cdot p_{2}) +(x_{3} \cdot p_{3})+(x_{4} \cdot p_{4})+(x_{5} \cdot p_{5}) + (x_{6} \cdot p_{6})

    Since the probability of each outcome is same

    E(X) = p(x_{1}+ x_{2}+ x_{3}+ x_{4}+ x_{5}+x_{6}) \\E(X) = \frac{1}{6} (-4 -3 -2 -1+0+1)\\E(X) = \frac{1}{6} (-9)\\E(X) = \frac{-9}{6} \\E(X) = -1.5

    Therefore, we can conclude that this game is not fair since the expected value is negative which represents a loss rather than a gain.

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45:7+7-4:2-5:5*4+35:2 =? ( )