Answer:
Case 1 matches with [tex]\frac{5}{48}[/tex].
Case 2 matches with [tex]\frac{1}{12}[/tex].
Case 3 matches with [tex]\frac{1}{4}[/tex].
Case 4 matches with [tex]\frac{1}{8}[/tex].
Case 5 matches with [tex]\frac{5}{16}[/tex].
Step-by-step explanation:
We have a total sample space of six-sided die and an eight sided die.
(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(1,7),(1,8)
(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(2,7),(2,8)
(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(3,7),(3,8)
(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(4,7),(4,8)
(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(5,7),(5,8)
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6),(6,7),(6,8)
Case 1: The Probability that both numbers are odd numbers and their product is greater than 10. are: (3,5),(3,7),(5,3),(5,5),(5,7)
And total number of outcomes are: 48.
So, required probability is:[tex]\frac{5}{48}[/tex].
Case 2: The probability that second number is twice the first number are:
(1,2),(2,4),(3,6),(4,8)
And total number of outcomes are: 48.
So, required probability is:[tex]\frac{4}{48}=\frac{1}{12}[/tex].
Case 3: The probability of a number whose sum is multiple of 4 are:
(1,3),(1,7),(2,2),(2,6),(3,1),(3,5),(4,4),(4,8),(5,3),(5,7),(6,2),(6,6).
And total number of outcomes are: 48.
So, required probability is:[tex]\frac{12}{48}=\frac{1}{4}[/tex].
Case 4: The probability that sum of two numbers is greater than 11.
(4,8),(5,8),(6,7),(6,8),(5,7),(6,6)
And total number of outcomes are: 48.
So, required probability is:[tex]\frac{6}{48}=\frac{1}{8}[/tex].
Case 5:The probability that second number rolled is less than first number are:
(2,1),(3,1),(3,2),(4,1),(4,2),(4,3),(5,1),(5,2),(5,3),(5,4),(6,1),(6,2),(6,3),(6,4),(6,5)
And total number of outcomes are: 48.
So, required probability is:[tex]\frac{15}{48}=\frac{5}{16}[/tex].
So, the correct matches are:
Case 1 matches with [tex]\frac{5}{48}[/tex].
Case 2 matches with [tex]\frac{1}{12}[/tex].
Case 3 matches with [tex]\frac{1}{4}[/tex].
Case 4 matches with [tex]\frac{1}{8}[/tex].
Case 5 matches with [tex]\frac{5}{16}[/tex].
