Example 1

A cubical die and a tetrahedral die are thrown together.

a) If X is the discrete random variable “total scored”, write down the probability distribution for X.

b) Find E(X).

c) Find Var(X).

A game is now played with the two dice. Anna has the cubical die and Beth has the tetrahedral die. They each gain points according to the following rules:

• If the number on both dice is greater than 3, then Beth gets 6 points.

• If the tetrahedral die shows 3 and the cubical die less than or equal to 3, then Beth gets 4 points.

• If the tetrahedral die shows 4 and the cubical die less than or equal to 3, then Beth gets 2 points.

• If the tetrahedral die shows a number less than 3 and the cubical die shows a 3, then Anna gets 5 points.

• If the tetrahedral die shows a number less than 3 and the cubical die shows a 1 or a 2, then Anna gets 3 points.

• If the tetrahedral die shows 3 and the cubical die greater than 3, then Anna gets 2 points.

• If the tetrahedral die shows a number less than 3 and the cubical die shows a number greater than 3, then Anna gets 1 point.

d) Write out the probability distribution for Y, “the number of points gained by Anna”.

e) Calculate E(Y) and Var(Y).

f) The game is now to be made fair by changing the number of points Anna gets when the tetrahedral die shows a number less than 3 and the cubical die shows a 1 or a 2. What is this number of points to the nearest whole number?

a) A probability space diagram is the easiest way to show the possible outcomes.

Hence the probability distribution for X is:

x
P(X=x)

b) Since the probability distribution is symmetrical E(X)=6.

c)

d) By considering the possibility space diagram again the probability distribution is:

y	-6	-4	-2
P(Y=y)

f) Let the number of points Anna gains when the tetrahedral die shows a number less than 3 and the cubical die shows a 1 or a 2 be y.

In this case the probability distribution is now:

y	-6	-4	-2		y
P(Y=y)

Since the game is now fair,