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December 19, 2014

December 19, 2014

Posted by **Emily** on Monday, December 15, 2008 at 10:12pm.

-- Here, I thought you could to 6! to equal 720 possible combinations, but my teacher marked this problem wrong.

14. Two dice are rolled. What is the probability that either the sum of the numbers showing is 7 or both numbers are even?

-- I do not understand how to do this problem at all :(

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Can someone help me with these problems? Any help is greatly appreciated :) thanks!!

- Algebra II -
**Anonymous**, Monday, December 15, 2008 at 11:14pm7) 6!/3!= 120

- Algebra II -
**Reiny**, Monday, December 15, 2008 at 11:24pmThere are 6 ways to get a 7

there are 9 ways to have both even

these are mutually exclusive events, that is, you can't have a sum of 7 AND both be even

so Prob(a sum of seven OR both are even)

= 16/36 = 4/9

or by formula

Prob(sum of seven OR both even)

= Prob(sum of seven) + prob(both even) - prob(sum of seven AND both even)

= 6/36 + 9/36 - 0

= 4/9

- Algebra II -
**Emily**, Monday, December 15, 2008 at 11:33pmOoooohhhhh I totally get it now!! Thanks anonymous and Reiny :)

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