Question

two dice are rolled. let x and y denote, respectively, the largest and smallest values obtained. compute the conditional mass function of y given x

Answer 1

Combinations mentioned among 36 possibilities.

Probability

A probability is a number that reflects the chance or likelihood that a particular event will occur. Probabilities can be expressed as proportions that range from 0 to 1, and they can also be expressed as percentages ranging from 0% to 100%.

The probability distribution of X is:

$P(X = x) = \frac{1}{6}$

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

Same for Y:

$P(Y=y) = \frac{1}{6}$

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 variables (X,Y). This probability is given by:

$P[ (X , Y) = (x , y) ] = \frac{1}{36}$

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 $P (Z = 2 ) = \frac{1}{36}$ , 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)  obtain $P(Z= 4) = \frac{3}{36}$  this is because  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) obtain $P(Z= 4) = \frac{3}{36}$  because we have 3 possibilities to obtain any of the combinations mentioned among 36 possibilities.

Combinations mentioned among 36 possibilities

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