Concept explainers
(a)
To find:The elements in the
(a)
Answer to Problem 8WE
The elements in the sample space are 216.
Explanation of Solution
Given:
A red, a blue and a green die are tossed.
Calculation:
Sample space is the set of the possible outcomes.
Calculate the elements,
Thus, the elements in the sample space are 216.
(b)
To find:The event that all three dice show the same number.
(b)
Answer to Problem 8WE
The event that all three dice show the same number is {(1,1,1),(2,2,2),(3,3,3),(4,4,4),(5,5,5),(6,6,6)}.
Explanation of Solution
Given:
A red, a blue and a green die are tossed.
Calculation:
Sample space is the set of the possible outcomes.
The required event for the given condition is {(1,1,1),(2,2,2),(3,3,3),(4,4,4),(5,5,5),(6,6,6)}
Thus, the event that all three dice show the same number is {(1,1,1),(2,2,2),(3,3,3),(4,4,4),(5,5,5),(6,6,6)}.
(c)
To find:The event that the sum of the numbers showing on the red and blue dice is less than the number showing on the green die.
(c)
Answer to Problem 8WE
The event that the sum of the numbers showing on the red and blue dice is less than the number showing on the green die is {(1,2,3),(1,1,4),(1,1,5),(1,1,6),(1,2,4),(1,2,5),(1,2,6),(2,1,4),(2,1,5),(2,1,6),(2,2,5),(2,2,6),(2,3,6),(3,1,5),(3,1,6),(3,2,6),(4,1,6)}.
Explanation of Solution
Given:
A red, a blue and a green die are tossed.
Calculation:
Sample space is the set of the possible outcomes.
The required event for the given condition is {(1,2,3),(1,1,4),(1,1,5),(1,1,6),(1,2,4),(1,2,5),(1,2,6),(2,1,4),(2,1,5),(2,1,6),(2,2,5),(2,2,6),(2,3,6),(3,1,5),(3,1,6),(3,2,6),(4,1,6)}
Thus, the event that the sum of the numbers showing on the red and blue dice is less than the number showing on the green die.is {(1,2,3),(1,1,4),(1,1,5),(1,1,6),(1,2,4),(1,2,5),(1,2,6),(2,1,4),(2,1,5),(2,1,6),(2,2,5),(2,2,6),(2,3,6),(3,1,5),(3,1,6),(3,2,6),(4,1,6)}
Chapter 15 Solutions
Algebra and Trigonometry: Structure and Method, Book 2
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