If you win $100 for rolling a 12, win $10 for rolling a sum less than 5, $5 for rolling a 5, and lose $3 for rolling anything else, what is the expected value of your winnings (or losses)

I suppose that the event here is rolling two dice.

The expected value can be written as.

EV = ∑xₙ*pₙ

Where x is the event and p is the probability of that event, that will be equal to the quotient between the number of outcomes that meet the given event and the total number of outcomes.

First, rolling two dice has a sample space equal to the product of the possible outcomes for each dice.

Each dice has 6 possible outcomes

Then the possible outcomes of two of them is:

6*6 = 36

x₁ = +$100,

p₁ = there is only on outcome that adds up to 12, then p₁ = 1/36.

x₂ = +$10

p₂ is the probability of rolling a sum less than 5.

The outcomes are:

1 and 1

1 and 2

2 and 1

2 and 2

So we have 4 possible outcomes, p₂ = 4/36.

x₃ = +5

p₃ is the probability of rolling a 5.

there are two outcomes:

2 + 3

3 + 2

then: p₃ = 2/36.

x₄ = -$3.

p₄ is conformed by all the other outcomes, we already have 1 + 4 + 2 = 7 used, then p₄ = (36 - 7)/36 = 29/36

Now we can write the expected value as:

EV = (1/36)*($100 + 4*$5 + 2*$3 - 29*$3) = $1.083

answer: 4

step-by-step explanation:

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204