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December 17, 2014

December 17, 2014

Posted by **Diana** on Tuesday, February 15, 2011 at 8:35am.

- Math -
**bobpursley**, Tuesday, February 15, 2011 at 8:46ameven sum 2, 4, 6, 8 , 10, 12

ways 1,1 one way

3,1;2,2; 1,3 three ways

5,1;4,2;3,3;3,3;1,5;2;4 six ways

and continue...

So for the pr (2)=1/6*1/6

pr(4)=1/6*1/6*3

pr(6)=1/6*1/6*6

and continue

- Math -
**Reiny**, Tuesday, February 15, 2011 at 9:11amTo have an even sum

1. all four must be even (E)

2. all four must be odd (O)

3 two must be even, two must be odd.

prob that one die lands even = 3/6 = 1/2

the same as for odd.

prob of EEEE = (1/2)^4

prob of OOOO = (1/2)^4

prob of EEOO = 4!/(2!2!) (1/2)^4

sum = 1/16 + 6(1/16) + 1/16 = 8/16 = 1/2

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