Dice. When two balanced dice are rolled, 36 equally likely outcomes are possible, as depicted in Fig. 5.1 on page 198. Let Ydenote the sum of the dice.

(a) What are the possible values of the random variable Y?

(b) Use random-variable notation to represent the event that the sum of the dice is 7.

(c) Find $P(Y=7)$.

(d) Find the probability distribution of Y. Leave your probabilities in fraction form.

e. Construct a probability histogram for Y.

In the game of craps, a first roll of a sum of 7 or 11 wins, whereas a first roll of a sum of 2, 3, or 12 loses. To win with any other first sum, that sum must be repeated before a sum of 7 is rolled. Determine the probability of

(f) a win on the first roll.

(g) a loss on the first roll.

The random variable is the sum of the outcomes from the two balanced dice and there are equally likely outcomes. The below table gives all possible outcomes.

A die can take on the values from 1 to 6.

The sum of two dice can then be at least 2 and at most 12.

Hence the possible outcomes are:

$2,3,4,5,6,7,8,9,10,11,12$

The event that the sum of two dice Y is 7 can be notated as:

$\left\{Y=7\right\}$

In the above table, it can be noted that 6 of the 36 outcomes result in 7.

Hence,

$$\begin{gathered} P(Y=7)=6/36 \\ =1/6 \end{gathered}$$

From the above-given table,

$$\begin{gathered} P\left(Y=2\right)=\frac{1}{36} \\ P\left(Y=3\right)=\frac{2}{36} \\ =\frac{1}{18} \\ P\left(Y=4\right)=\frac{3}{36} \\ =\frac{1}{12} \\ P\left(Y=5\right)=\frac{4}{36} \\ =\frac{1}{9} \\ P\left(Y=6\right)=\frac{5}{36} \\ P\left(Y=7\right)=\frac{6}{36} \\ =\frac{1}{6} \\ P\left(Y=8\right)=\frac{5}{36} \\ P\left(Y=9\right)=\frac{4}{36} \\ =\frac{1}{9} \\ P\left(Y=10\right)=\frac{3}{36} \\ =\frac{1}{12} \\ P\left(Y=11\right)=\frac{2}{36} \\ =\frac{1}{18} \\ P\left(Y=12\right)=\frac{1}{36} \end{gathered}$$

Therefore the below table gives the probability distribution

The height of the bars is equal to the probability and the bars are centered on the possible values of Y.

$$\begin{gathered} P\left(win\right)=P(Y=7)+P(Y=11) \\ =\frac{6}{36}+\frac{2}{36} \\ =\frac{2}{9} \end{gathered}$$

$$\begin{gathered} P\left(loss\right)=P(Y=2)+P(Y=3)+P(Y=12) \\ =\frac{1}{36}+\frac{2}{36}+\frac{1}{36} \\ =\frac{1}{9} \end{gathered}$$