Question 1097231
Two dice are rolled. 
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Here are all possible rolls:  Count them, there are 36

(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)

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Find the probability of getting the following.
2) a sum of 6 or 7 or 8 
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I'll color those red:

(1,1) (1,2) (1,3) (1,4) <font color="red"><b>(1,5) (1,6)</font></b>

(2,1) (2,2) (2,3) <font color="red"><b>(2,4) (2,5) (2,6)</font></b>

(3,1) (3,2) <font color="red"><b>(3,3) (3,4) (3,5)</font></b> (3,6) 

(4,1) <font color="red"><b>(4,2) (4,3) (4,4)</font></b> (4,5) (4,6) 

<font color="red"><b>(5,1) (5,2) (5,3)</font></b> (5,4) (5,5) (5,6)

<font color="red"><b>(6,1) (6,2)</font></b> (6,3) (6,4) (6,5) (6,6)

Count the red ones.  I count 16.  So the probability
of rolling one on those 16, is 16 times out of 36,
or 16/36 which reduces to 4/9.

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b) doubles or a sum of 4 or 6 
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I'll color those red:

<font color="red"><b>(1,1)</font></b> (1,2) <font color="red"><b>(1,3)</font></b> (1,4) <font color="red"><b>(1,5)</font></b> (1,6)

(2,1) <font color="red"><b>(2,2)</font></b> (2,3) <font color="red"><b>(2,4)</font></b> (2,5) (2,6)

<font color="red"><b>(3,1)</font></b> (3,2) <font color="red"><b>(3,3)</font></b> (3,4) (3,5) (3,6) 

(4,1) <font color="red"><b>(4,2)</font></b> (4,3) <font color="red"><b>(4,4)</font></b> (4,5) (4,6) 

<font color="red"><b>(5,1)</font></b> (5,2) (5,3) (5,4) <font color="red"><b>(5,5)</font></b> (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) <font color="red"><b>(6,6)</font></b>

Count the red ones.  I count 12.  So the probability
of rolling one on those 12, is 12 times out of 36,
or 12/36 which reduces to 1/3.

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c) a sum greater than 9 or less than 4,
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Now you can do that one by yourself, right?

Edwin</pre>