Suppose someone gives you 8 to 2 odds that you cannot roll two even numbers with the roll of two fair dice. This means you win $ if you succeed and you lose $ if you fail. What is the expected value of this game to you? Should you expect to win or lose the expected value in the first game? What can you expect if you play times? Explain.

Complete question :

Suppose someone gives you 8 to 2 odds that you cannot roll two even numbers with the roll of two fair dice. This means you win $8 if you succeed and you lose $2 if you fail. What is the expected value of this game to you? What can you expect if you play 100 times.

$0.5 ; win $50 with 100 rolls

Step-by-step explanation:

From a roll of two fair dice; probability of obtaining an even number :

Even numbers = (2, 4, 6) = 3

P = 3 /6 = 1 /2

For 2 fair dice ; probability of rolling two even numbers : independent event.

1/2 * 1/2 = 1/4

Hence, p(success) = 1/4 ; P(failure) = 1 - 1/4 = 3/4

Probability table

Winning = $8 or loss = - $2

X : 8 - 2

P(x) __ 1/4 3/4

Expected value : E(x) = ΣX*P(x)

E(x) = (8 * 1/4) + (-2 * 3/4)

E(x) = 2 - 1.5

E(x) = $0.5

Since expected value is positive, the expect to win

If played 100 times;

Expected value = 100 * $0.5 = $50

the ratio of balloons to tables is 5: 2 and the constant variation is 5.

the equation is y=5x

final total: 63

step-by-step explanation:

60 times sales tax

60 times 0.05

3

60 + 3

63