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?

Answer :

Answer:

[tex]E(X) = -1.5[/tex]

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

[tex]E(X) = \sum(x_{i}p_{i})[/tex]

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

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

[tex]p = \frac{1}{6}[/tex]

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

[tex]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\\[/tex]

Now we can find the expected value of this game,[tex]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})[/tex]

Since the probability of each outcome is same

[tex]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[/tex]

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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