Question

Suppose you roll a fair six-sided die. Then you roll a fair eight-sided die. Match each probability to its correct value.

Answer 1

Answer:

A) matches $\frac{5}{48}$

B) matches $\frac{1}{12}$

C) matches $\frac{1}{4}$

D) matches $\frac{1}{8}$

E) matches $\frac{5}{16}$

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  $=\frac{Number Of Ways Event Can Occur}{Sample Space}$

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:

                                 $=\frac{5}{48}$

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:                  

                       $=\frac{4}{48}$  $=\frac{1}{12}$

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:

                       $=\frac{12}{48}$  $=\frac{1}{4}$

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:

                      $=\frac{6}{48}$  $=\frac{1}{8}$

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:

                 $=\frac{15}{48}$  $=\frac{5}{16}$