Question

You roll a standard die. You win $12 if you roll a six, 56 if you roll a one, and $0 if you roll anything else. What is the expected value of a single roll?
$2

$18

$59

$3​

Answer 1

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:

brainly.com/question/22097128