Concept explainers
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
(a)
To find: The value that could be taken by the random variable x.
Answer to Problem 17CR
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 |
(b)
The probability distribution of the variable x.
Answer to Problem 17CR
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:
The probability of getting sum 2 can be calculated as:
The probability of getting sum 3 can be calculated as:
The probability of getting sum 4 can be calculated as:
The probability of getting sum 5 can be calculated as:
The probability of getting sum 6 can be calculated as:
The probability of getting sum 7 can be calculated as:
The probability of getting sum 8 can be calculated as:
The probability of getting sum 9 can be calculated as:
The probability of getting sum 10 can be calculated as:
The probability of getting sum 11 can be calculated as:
The probability of getting sum 12 can be calculated as:
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 |
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Chapter 5 Solutions
Understanding Basic Statistics
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