Question

In this experiment, both a fair four-sided die and a fair six-sided die are rolled (these dice both have the numbers most people would expect on them). Let Z be a random variable that represents the absolute value of their difference. For instance, if a 4 and a 1 are rolled, the corresponding value of Z is 3.

(a) What is the pmf of Z?
(b) Draw a graph of the cdf of Z

Answer 1

Let (x,y) represent the result obtained on roling the 2 dice. x is the number rolled on the 4-sided die and y be the number rolled on the 6-sided die.

Total number of possible (x,y) pairs = 4 * 6 = 24 since x can ntake 4 values and y can take 6 values.

The possible values of Z are : {0 , 1 , 2 , 3 , 4 , 5}

a. To find the pmf of Z:

Z = 0, when the result is one of the following: (1,1), (2,2), (3,3), (4,4)

Z = 1, when the result is one of the following: (1,2), (2,3), (3,4), (4,5), (2,1), (3,2), (4,3)

Z = 2, when the result is one of the following: (1,3), (2,4), (3,5), (4,6), (3,1), (4,2)

Z = 3, when the result is one of the following: (1,4), (2,5), (3,6), (4,1)

Z = 4, when the result is one of the following: (1,5), (2,6)

Z = 5, when the result is one of the following: (1,6)

Therefore the pmf of Z is given by:

7 24

$P(Z = 2) = \frac{6}{24} = \frac{1}{4}$

$P(Z = 3) = \frac{4}{24} = \frac{1}{6}$

$P(Z = 4) = \frac{2}{24} = \frac{1}{12}$

$P(Z = 5) = \frac{1}{24}$

b. To find the cdf of Z:

Let F(z) be the cdf of Z. Then

F(z) = P(Z < z)

Therefore,

$F(0) = P(Z \leqslant 0) = \frac{4}{24}$

$F(1) = P(Z \leqslant 1) = P(Z = 0) + P(Z = 1) = \frac{11}{24}$

$F(2) = P(Z \leqslant 2) = P(Z = 0) + P(Z = 1) + P(Z = 2) = \frac{17}{24}$

$F(3) = P(Z \leqslant 3) = P(Z = 0) + P(Z = 1) + P(Z = 2) + P(Z = 3) = \frac{21}{24}$

$F(4) = P(Z \leqslant 4)$

  $= P(Z = 0) + P(Z = 1) + P(Z = 2) + P(Z = 3) + P(Z = 4) = \frac{23}{24}$

$F(5) = P(Z \leqslant 5)$

  

Graph of cdf of Z: