SOLUTION: An ordinary (fair) die is a cube with the numbers 1 through 6 on the sides (represented by painted spots). Imagine that such a die is rolled twice in succession and that the face v
Algebra.Com
Question 1117343: An ordinary (fair) die is a cube with the numbers 1 through 6 on the sides (represented by painted spots). Imagine that such a die is rolled twice in succession and that the face values of the two rolls are added together. This sum is recorded as the outcome of a single trial of a random experiment.
Compute the probability of each of the following events:
Event : The sum is greater than 7 .
Event : The sum is not divisible by 3 and not divisible by 5.
Write your answers as exact fractions.
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
An ordinary (fair) die is a cube with the numbers 1 through 6 on the sides (represented by painted spots). Imagine that such a die is rolled twice in succession and that the face values of the two rolls are added together. This sum is recorded as the outcome of a single trial of a random experiment.
Compute the probability of each of the following events:
----
You can figure out these answers.
Sketch a 6 by 6 square.
Label the rows as 1,2,3,4,5,6
Label the columns as 1,2,3,4,5,6
Note:: There will be 6*6 = 36 entries In the square.
1st row sums:: 2,3,4,5,6,7
2nd row sums:: 3,4,5,6,7,8
etc.
Finish the square
bottom row sums:: 7,8,9,10,11,12
-----------------------------
Event : The sum is greater than 7 .
Event : The sum is not divisible by 3 and not divisible by 5.
Write your answers as exact fractions.
-----
Cheers,
Stan H.
-----------
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
There are 36 possible outcomes:
One way to make 2, two ways to make 3, ... 6 ways to make 7, 5 ways to make 8, and so on.
The results greater than 7 include 8, 9, 10, 11, and 12, all together, 5 plus 4 plus 3 plus 2 plus 1 = 15. So there are 15 ways to get something larger than 7 out of 36 total outcomes.
2, 4, 5, 7, 8, 10, and 11 are not divisible by 3. Of these, 5 and 10 are not divisible by 5. 1 way to make 2, 3 ways to make 4, 6 ways to make 7, 5 ways to make 8. 2 + 3 + 6 + 5 = 16 successful outcomes out of 36 total outcomes.
I'm going to presume that you know how to write a fraction and reduce it to lowest terms
John
My calculator said it, I believe it, that settles it
RELATED QUESTIONS
An ordinary (fair) die is a cube with the numbers
1
through
6
on the sides... (answered by solver91311)
An ordinary (fair) die is a cube with the numbers
1 through 6 on the sides (represented (answered by Edwin McCravy)
An ordinary (fair) die is a cube with the numbers
1
through
6
on the sides... (answered by Boreal)
An ordinary (fair) die is a cube with the numbers
1
through
6
on the sides... (answered by ikleyn)
An ordinary (fair) die is a cube with the numbers 1 through 6 on the sides (represented... (answered by stanbon)
An ordinary (fair) die is a cube with the numbers 1 through 6 on the sides (represented... (answered by Boreal)
An ordinary (fair) die is a cube with the numbers 1 through 6 on the sides (represented... (answered by ikleyn)
Die rolling
An ordinary (fair) die is a cube with the numbers through on the sides... (answered by edjones)
An ordinary (fair) die is a cube with the numbers through on the sides (represented by... (answered by Boreal)
An ordinary (fair) die is a cube with the numbers through on the sides (represented by... (answered by stanbon)