Suppose you roll a fair six-sided die. Then you roll a fair eight-sided die. Match each probability to its correct value. A) the probability that both numbers are odd numbers and their product is greater than 10 B) the probability that the second number is twice the first number C) the probability of getting numbers whose sum is a multiple of 4 D) the probability that the sum of the two numbers is greater than 11 E) the probability that the second number rolled is less than the first number 5/16 arrowRight 1/8 arrowRight 1/12 arrowRight 1/4 arrowRight 5/48 arrowRight
Accepted Solution
A:
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) = 48Probability [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 = 5Sample space = 48The 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 = 4Sample space = 48The 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 = 12Sample space = 48The 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 = 6Sample space = 48the 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 = 15Sample space = 48The probability that the second number rolled is less than the first number: [tex]=\frac{15}{48}[/tex] [tex]=\frac{5}{16}[/tex]