IDNLearn.com is your go-to resource for finding expert answers and community support. Our experts are available to provide accurate, comprehensive answers to help you make informed decisions about any topic or issue you encounter.
Sagot :
Answer:
The probability that the value of the second die is higher than the first
[tex]P(E) = \frac{15}{36} = \frac{5}{12}[/tex]
Step-by-step explanation:
Explanation:-
In a single throw with two dice total number of sums = 6 X 6 = 36
Let 'E' be the event of the second die is higher than the first
Total number of cases
= {( 1,1), (1,2 )( 1,3), (1,4) , (1,5) ,(1,6)
( 2,1), (2,2 )( 2,3), (2,4) , (2,5),(2,6)
( 3,1), (3,2 )( 3,3), (3,4) , (3,5),(3,6)
( 4,1), (4,2 )( 4,3), (4,4) , (4,5),(4,6)
( 5,1), (5,2 )( 5,3), (5,4) , (5,5),(5,6)
( 6,1), (6,2 )( 6,3), (6,4) , (6,5),(6,6)}
The favourable cases
n(E) = {(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6),(5,6)} = 15
The probability that the value of the second die is higher than the first
[tex]P(E) = \frac{15}{36} = \frac{5}{12}[/tex]
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. IDNLearn.com has the solutions to your questions. Thanks for stopping by, and come back for more insightful information.
