SOLUTION: 1.1 A fair die is rolled twice. Find the probability of the following events: a. The second number is twice the first b. The second number is not greater than the first c. At

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Question 1145654: 1.1 A fair die is rolled twice. Find the probability of the following events:
a. The second number is twice the first
b. The second number is not greater than the first
c. At least one number is greater than
3. 1.2 Two distinct diceAand Barerolled. Whatis the probabilityofeachofthe following events?
a. At least one 4 appears
b. Just one 4 appears
c. The sum of the face values is 7
d. One of the values is 3 and the sum of the two values is 5
e. One of the values is 3 or the sum of the two values is 5
Found 2 solutions by ikleyn, Edwin McCravy:
Answer by ikleyn(52832) About Me  (Show Source):
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(a)  In all, there are 6*6 = 36 possible outcomes, with equal probability %281%2F6%29%2A%281%2F6%29 = 1%2F36 each.


     The favorable outcomes are (1,2), (2,4), and (3,6). In all, there are 3 favorable outcomes.


      Therefore, the answer is  3%2F36 = 1%2F12.   



(b)  The favorable outcomes are (1,1), 

                                (2,1), (2,2)

                                (3,1), (3,2), (3,3)

                                . . . . . . . . . . . . 

                                (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)


     In all, there are  %286%2A5%29%2F2 + 6 = 15 + 6 = 21 favorable outcomes.


     Therefore, the answer in this case is  21%2F36 = 7%2F12.


(c)  What your "c" means, is totally unclear.

------------------

If you want to learn this subject,  look into the lesson
    - Rolling a pair of fair dice
in this site.

After reading it,  you will become an expert in the area.



Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
1.1 A fair die is rolled twice. 
Here is the sample space of the 36 possible pairs of rolls:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
 
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
 
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6) 
 
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6) 
 
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
 
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
a. The second number is twice the first 
I'll color those red  

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
 
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
 
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6) 
 
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6) 
 
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
 
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Count them.  There are 3.  The probability is 3 out of 36 or 3/36 
which reduces to 1/12.
b. The second number is not greater than the first 
I'll color those red:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
 
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
 
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6) 
 
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6) 
 
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
 
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Count them.  There are 21.  The probability is 21 out of 36 or 21/36 
which reduces to 7/12.
c. At least one number is greater than 3. 
I'll color those red:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
 
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
 
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6) 
 
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6) 
 
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
 
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Count them.  There are 27.  The probability is 27 out of 36 or 27/36 
which reduces to 3/4.
1.2 Two distinct dice A and B are rolled. 
That's the same as rolling one die twice. So the sample space is the same.

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
 
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
 
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6) 
 
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6) 
 
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
 
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
What is the probability of each of the following events?
a. At least one 4 appears 
I'll color those red.

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
 
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
 
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6) 
 
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6) 
 
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
 
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Count them.  There are 11.  The probability is 11 out of 36 or 11/36 
which does not reduce.
b. Just one 4 appears 
I'll color those red:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
 
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
 
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6) 
 
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6) 
 
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
 
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Count them.  There are 10.  The probability is 10 out of 36 or 10/36 
which reduces to 5/18.
c. The sum of the face values is 7 
I'll color those red:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
 
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
 
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6) 
 
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6) 
 
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
 
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Count them.  There are 6.  The probability is 6 out of 36 or 6/36 
which reduces to 1/6.
d. One of the values is 3 and the sum of the two values is 5. 
I'll color those red:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
 
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
 
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6) 
 
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6) 
 
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
 
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Count them.  There are 2.  The probability is 2 out of 36 or 1/18 
which reduces to 1/12.
e. One of the values is 3 or the sum of the two values is 5 
I'll color those red.

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
 
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
 
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6) 
 
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6) 
 
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
 
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Count them.  There are 13.  The probability is 13 out of 36 or 13/36 
which does not reduce.

Edwin