contestada

You roll a die. if it comes up a 6, you win $100. if not, you get to roll again. if you get a 6 the second time, you win $50. if not, you lose.

a.create a probability model for the amount you win.

b.find the expected amount you'll win.

c.what would you be willing to pay to play this game?

Respuesta :

jmonterrozar

Answer:

a.

x $         0 $         50         $ 100

p (x)    25/36      5/36        6/36

b. 23.61%

c. 23.61%

Step-by-step explanation:

We have that "x" is the amount of money earned:

we have to fail twice, the probability of failure is 5/6 and the probability of hitting is 1/6, therefore if you lose the probability would be:

5/6 * 5/6 = 25/36

the one to win 50 $, would be:

1/6 * 5/6 = 5/36

and the one to win $ 100 would be:

1/6 * = 6/36

that is to say:

a.

x $         0 $         50         $ 100

p (x)    25/36      5/36        6/36

b. the expected amount you'll win would be:

0 * 25/36 + 50 * 5/36 + 100 * 6/36 = 23.61

c. The normal thing is to pay at least the expected amount, that is, $ 23.61, since something greater than that would already be profit.

fichoh

The expected value is the sum of the product of values and their respective probabilities. Hence ;

  • Expected winning = $9.722

  • Willing amount to be paid should be any amount less than $9.722

To win $100 :

  • P(6) × P(6) = 1/6 × 1/6 = 1/36

To win $50 :

  • P(not 6) × P(6) = 5/6 × 1/6 = 5/36

To win nothing :

  • P(not 6) × P(not 6) = 5/6 × 5/6 = 25/36

The probability distribution table :

  • X : _______ $0 ______ $50 _______ $100

  • P(X) : ____ 25/36 _____ 5/36 _______ 1/36

The expected value :

  • E(X) = Σ[(X × P(X)]

E(X) = (0 × 25/36) + (50 × 5/36) + (100 × 1/36)

E(X) = 0 + 9.7222

E(X) = 9.722

Therefore, the expected amount to be won is $9.722

Willing amount to be paid to play the game :

Any value below the expected value,

Therefore, the willing amount will be < 9.722

Learn more :https://brainly.com/question/18405415

Otras preguntas