Question

Two dice are rolled. Let the random variable X denote the number that falls uppermost on the first die and let Y denote the number that falls uppermost on the second die.

(a) Find the probability distributions of X and Y.
(b) Find the probability distribution of X + Y.

Answer 1

Explanation:

a) Lets start by giving the answer:

The probability distribution of X is:

`P(X=x)=(1)/(6)`

For every x = 1, 2, 3, 4, 5, 6

Same for Y:

`P(Y=y)=(1)/(6)`

For every y = 1, 2, 3, 4, 5, 6

This is because each variable has 6 possibilities which are equiprobable. In other words, every number (1 to 6) has the same probability of falling uppermost on each dice.

Now if the question refers to a joint probability distribution, then we have a joint random viariable (X,Y). This probability is given by:

`P[(X,Y)=(x,y)] = (1)/(36)`

For every x = 1, 2, 3, 4, 5, 6 and y = 1, 2, 3, 4, 5, 6

The reason for this is that the combination of the result of both dices produces 36 equiprobable possibilities.

b) For the probability distribution of the sum of the variables we may define:

`Z=X+Y`

The for the random variable Z the next results are possible:

Z P(Z)

2 1/36

3 2/36

4 3/36

5 4/36

6 5/36

7 6/36

8 5/36

9 4/36

10 3/36

11 2/36

12 1/36

This is our probability distribution for the sum of X+Y. Now to understand the results lets see some examples:

For z = 2 (x=1 and y=1) we obtain the probability
`P(Z=2)=(1)/(36)`, this is because we have 1 possibility to obtain (1,1) among 36 possibilities.

For z = 4 that is for (x=1, y=3) and (x=2, y=2) and (x=3, y=1) we obtain
`P(Z=4)=(3)/(36)` this is because we have 3 possibilities to obtain any of the combinations mentioned among 36 possibilities.

For z = 10 that is for (x=4, y=6) and (x=5, y=5) and (x=6, y=6) we obtain
`P(Z=4)=(3)/(36)` because we have 3 possibilities to obtain any of the combinations mentioned among 36 possibilities.