MATH SOLVE

5 months ago

Q:
# For a special game Don has two 8-sided fair dice numbered from 1 to 8 on the faces. Just as with ordinary dice, Don rolls the dice and sums the numbers that appear on the top face, getting a sum from 2 to 16. What is the sum of the possible numbers which have a probability of 1/32 of appearing

Accepted Solution

A:

AnswerThe sum of the possible numbers are {(1,2), (2,1)}
Step-by-step explanation:Probability for rolling two dice with the eight sided dots are 1, 2, 3, 4, 5, 6, 7 and 8 dots in each die. When two dice are rolled or thrown simultaneously, thus number of event can be [tex]8^{2}[/tex]= 64 because each die has 1 to 8 numbers on its faces. Then the possible outcomes of the sample space are shown in the pdf document below . As Don rolls the dice and sums the number that appear on the top face, he gets a sum from 2 to 16.
Assuming; getting sum of 2 Let E[tex]_{1}[/tex] = event of getting sum of 2
E[tex]_{1}[/tex] = { (1 , 1 ) }
Therefore, Probability of getting sum of 2 will be; P ( E[tex]_{1}[/tex] ) = [tex]\frac{Number of favorable outcome}{Total number of possible outcome}[/tex]
= [tex]\frac{1}{64}[/tex]
GETTING SUM OF TWO ( i.e { (1 , 1 ) } ) will give the probability of 1/64 of appearing. But we are looking for the probability of 1/32 of appearing. Let look at the possibility of getting sum of 3.Assuming; getting sum of 3Let E[tex]_{2}[/tex] = event of getting sum of 3
E[tex]_{2}[/tex] = { (1 , 2 ) (2 , 1) }
Therefore, Probability of getting sum of 3 will be; P ( E[tex]_{2}[/tex] ) = [tex]\frac{Number of favorable outcome}{Total number of possible outcome}[/tex]
= [tex]\frac{2}{64}[/tex]
= [tex]\frac{1}{32}[/tex]
For all probability of getting the sum greater than 3 to 16 will be void because there wont be a chance for 1/32 to appear. Therefore, the sum of the possible numbers are: {(1,2), (2,1)} which have a probability of 1/32 of appearing.I hope this comes in handy at the rightful time!