You are going to your local pizza place for a slice, where the basic offering is

Question

You are going to your local pizza place for a slice, where the basic offering is plain cheese pizza. You can add extra toppings for 50 cents each... OR you can roll a pair of dice to determine how much you should pay for the extra topping. If you enter the game and roll the dice, the higher of the two resulting values will be multiplied by 10 cents, and the lowest one will simply add the corresponding number of cents. So, for example, if you roll a {1, 2}, your extra topping will cost you 21 cents. If you roll {2, 2), you pay 22 cents.

Let X represent the cost for your extra topping resulting from this game.

a. Is X a valid random variable? Elaborate.

b. Write a table listing all the possible outcomes of the random experiment and relate each of these with the numerical value to which it gets mapped though X.

c. Describe fully the probability distribution function (aka probability mass function) for X, and verify that it meets all the requirements of a distribution function.

d. Calculate the probability that by entering the game, you will end up paying more for your extra topping than you would have, had you chosen not to enter the game. (How do you feel about that?)

Explanation / Answer

a.

X is a valid random variable because all possible values of X are numerical outcomes of a random phenomenon rolling of a dice and it is mesurable.

b.

(c)

Probability of each outcome {a,b} = (1/6) *(1/6) = 1/36

So, PMF of X is given as,

Sum of all probabilities = (1/36) + (1/18)+ (1/18)+ (1/18)+ (1/18)+ (1/18)+ (1/36)+ (1/18)+ (1/18)+ (1/18)+ (1/18) + (1/36)+ (1/18)+ (1/18)+ (1/18) + (1/36)+ (1/18)+ (1/18) + (1/36)+ (1/18) + (1/36) = 1

As, each probability value is positive and Sum of all probabilities is 1, it meets all the requirements of a distribution function.

(d)

If we do not enter the game, we have to pay 50 cents. So the probability that by entering the game, we have to pay more than 50 cents is,

P(X > 50) = P(X = 51) + P(X = 61) + P(X = 52) + P(X = 62) + P(X = 53) + P(X = 63) + P(X = 54) + P(X = 64) + P(X = 55) + P(X = 65) + P(X = 66)

= (1/18)+ (1/18)+ (1/18)+ (1/18) + (1/18)+ (1/18)+ (1/18)+ (1/18) + (1/36)+ (1/18) + (1/36)

= 0.5556

Possible Outcomes X {1,1} 11 {1,2} {2,1} 21 {1,3} {3,1} 31 {1,4} {4,1} 41 {1,5} {5,1} 51 {1,6} {6,1} 61 {2,2} 22 {2,3} {3,2} 32 {2,4} {4,2) 42 {2,5} (5,2} 52 {2,6} {6,2} 62 {3,3} 33 {3,4} {4,3} 43 {3,5} {5,3} 53 {3,6} {6,3} 63 {4,4} 44 {4,5} {5,4} 54 {4,6} {6.4} 64 {5,5} 55 {5,6} {6,5} 65 {6,6} 66

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