Answer:
A) matches [tex]\frac{5}{48}[/tex]
B) matches [tex]\frac{1}{12}[/tex]
C) matches [tex]\frac{1}{4}[/tex]
D) matches [tex]\frac{1}{8}[/tex]
E) matches [tex]\frac{5}{16}[/tex]
Step-by-step explanation:
The sample space for the roll of six sided die and 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)
Therefore, the total number of outcomes (Sample space) = 48
Probability [tex]=\frac{Number Of Ways Event Can Occur}{Sample Space}[/tex]
A) The possible ways of obtaining both numbers are odd numbers and their product is greater than 10 are: (3,5), (3,7), (5,3), (5,5), (5,7)
Total number of possible ways = 5
Sample space = 48
The probability that both numbers are odd numbers and their product is greater than 10:
[tex]=\frac{5}{48}[/tex]
B) The possible ways of obtaining second number is twice the first number are: (1,2), (2,4), (3,6), (4,8)
Total number of possible ways = 4
Sample space = 48
The probability that the second number is twice the first number:
[tex]=\frac{4}{48}[/tex] [tex]=\frac{1}{12}[/tex]
C) The possible ways of getting numbers whose sum is a 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)
Total number of possible ways = 12
Sample space = 48
The probability of getting numbers whose sum is a multiple of 4:
[tex]=\frac{12}{48}[/tex] [tex]=\frac{1}{4}[/tex]
D) The possible ways of getting the sum of the two numbers is greater than 11 are: (4,8), (5,7), (5,8), (6,6), (6,7), (6,8)
Total number of possible ways = 6
Sample space = 48
the probability that the sum of the two numbers is greater than 11:
[tex]=\frac{6}{48}[/tex] [tex]=\frac{1}{8}[/tex]
E) The possible ways of getting the second number rolled is less than the first number are: (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)
Total number of possible ways = 15
Sample space = 48
The probability that the second number rolled is less than the first number:
[tex]=\frac{15}{48}[/tex] [tex]=\frac{5}{16}[/tex]
