Chapter 5, Problem 13CR

General: Two Dice In a game of craps, you roll two fair dice. Whether you win or lose depends on the sum of the numbers appearing on the tops of the dice. Let x be the random variable that represents the sum of the numbers on the tops of the dice.

(a) What values can x take on?

(b) What is the probability distribution of these x values (that is, what is the probability that x = 2. 3, etc.)?

To determine

To find: The value that could be taken by the random variable x.

Answer to Problem 13CR

Solution: The possible values that x can take are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12

Explanation of Solution

Given: Two dices are rolled and winning or losing the game depends on the sum of the numbers appearing on the dice.

Calculation: Consider x be the random variable, which represents the sum of the number that appears on the dice. Following are the values that variable x can take.

Sum	Numbers on 1st and 2nd die
2	1 and 1
3	1 and 2 or 2 and 1
4	1 and 3, 2 and 2 and 3 and 1
5	1 and 4, 2 and 3, 3 and 2, 4 and 1
6	1 and 5, 2 and 4, 3 and 3, 4 and 2, and 5 and 1
7	1 and 6, 2 and 5, 3 and 4, 4 and 3, 5, and 2 and 6 and 1
8	2 and 6, 3 and 5, 4 and 4, 5 and 3, and 6 and 2
9	3 and 6, 4 and 5, 5 and 4, 6 and 3
10	4 and 6, 5 and 5 and 6 and 4
11	5 and 6 or 6 and 5
12	6 and 6

To determine

The probability distribution of the variable x.

Answer to Problem 13CR

Solution: Following is the probability distribution of the values of the variable X.

x	2	3	4	5	6	7	8	9	10	11	12
P(x)	0.028	0.056	0.083	0.111	0.139	0.167	0.139	0.111	0.083	0.056	0.028

Explanation of Solution

Given: Two dices are rolled and variable x represents the sum of the number on the dices and consider the information provided in part (a) for the values that variable x can take.

Calculation: Consider outcomes are equally likely. The formula used to assign the probability based on equally likely outcomes is:

$\text{Probability of an event = }\frac{\text{Total number of outcomes favorable to event}}{\text{Total number of outcomes}}$

The probability of getting sum 2 can be calculated as:

$\begin{array}{c}\text{P}\left(Sum\text{ 2}\right)\text{ = }\frac{\text{Number of outcomes that show the sum 2}}{\text{Total number of outcomes}}\\ =\frac{1}{36}\\ =0.028\end{array}$

The probability of getting sum 3 can be calculated as:

$\begin{array}{c}\text{P}\left(Sum\text{ 3}\right)\text{ = }\frac{\text{Number of outcomes that show the sum 3}}{\text{Total number of outcomes}}\\ =\frac{2}{36}\\ =0.056\end{array}$

The probability of getting sum 4 can be calculated as:

$\begin{array}{c}\text{P}\left(Sum\text{ 4}\right)\text{ = }\frac{\text{Number of outcomes that show the sum 4}}{\text{Total number of outcomes}}\\ =\frac{3}{36}\\ =0.083\end{array}$

The probability of getting sum 5 can be calculated as:

$\begin{array}{c}\text{P}\left(Sum\text{ 5}\right)\text{ = }\frac{\text{Number of outcomes that show the sum 5}}{\text{Total number of outcomes}}\\ =\frac{4}{36}\\ =0.111\end{array}$

The probability of getting sum 6 can be calculated as:

$\begin{array}{c}\text{P}\left(Sum\text{ 6}\right)\text{ = }\frac{\text{Number of outcomes that show the sum 6}}{\text{Total number of outcomes}}\\ =\frac{5}{36}\\ =0.139\end{array}$

The probability of getting sum 7 can be calculated as:

$\begin{array}{c}\text{P}\left(Sum\text{ 7}\right)\text{ = }\frac{\text{Number of outcomes that show the sum 7}}{\text{Total number of outcomes}}\\ =\frac{6}{36}\\ =0.167\end{array}$

The probability of getting sum 8 can be calculated as:

$\begin{array}{c}\text{P}\left(Sum\text{ 8}\right)\text{ = }\frac{\text{Number of outcomes that show the sum 8}}{\text{Total number of outcomes}}\\ =\frac{5}{36}\\ =0.139\end{array}$

The probability of getting sum 9 can be calculated as:

$\begin{array}{c}\text{P}\left(Sum\text{ 9}\right)\text{ = }\frac{\text{Number of outcomes that show the sum 9}}{\text{Total number of outcomes}}\\ =\frac{4}{36}\\ =0.111\end{array}$

The probability of getting sum 10 can be calculated as:

$\begin{array}{c}\text{P}\left(Sum\text{ 10}\right)\text{ = }\frac{\text{Number of outcomes that show the sum 10}}{\text{Total number of outcomes}}\\ =\frac{3}{36}\\ =0.083\end{array}$

The probability of getting sum 11 can be calculated as:

$\begin{array}{c}\text{P}\left(Sum\text{ 11}\right)\text{ = }\frac{\text{Number of outcomes that show the sum 11}}{\text{Total number of outcomes}}\\ =\frac{2}{36}\\ =0.056\end{array}$

The probability of getting sum 12 can be calculated as:

$\begin{array}{c}\text{P}\left(Sum\text{ 12}\right)\text{ = }\frac{\text{Number of outcomes that show the sum 12}}{\text{Total number of outcomes}}\\ =\frac{1}{36}\\ =0.028\end{array}$

Therefore, the required probability distribution is:

x	2	3	4	5	6	7	8	9	10	11	12
P(x)	0.028	0.056	0.083	0.111	0.139	0.167	0.139	0.111	0.083	0.056	0.028