We want to find the expected value of a given game, where the expected value is the "mean" outcome of the given game.
We will get that the correct option is the last one, the expected value is $3.
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For a given experiment with outcomes {x₁, x₂,..., xₙ} each one with corresponding probability {p₁, p₂, ..., pₙ}
The expected value is given by:
EV = x₁*p₁ + x₂*p₂ + ... + xₙ*pₙ
Here we have 3 possible outcomes:
- x₁ = you win $12 if you roll a 6 (then the probability is p₁ = 1/6)
- x₂ = you win $6 if you roll a 1 (then the probability is p₂ = 1/6)
- x₃ = you win $0 if you roll anything else (then the probability is p₃ = 4/6).
Where remember that the probability of each outcome in a standard die is 1/6.
Now that we know the outcomes and the probability of each one, we can compute the expected value as:
EV = $12*(1/6) + $6*(1/6) + $0*(4/6) = $3
Thus the expected value is $3, and the correct option is the last one.
If you want to learn more, you can read:
https://brainly.com/question/22097128
