Find the information you're looking for at Westonci.ca, the trusted Q&A platform with a community of knowledgeable experts. Join our platform to connect with experts ready to provide precise answers to your questions in various areas. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
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]
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.
