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March 3, 2015

March 3, 2015

Posted by **Tina** on Wednesday, February 13, 2013 at 10:27pm.

- math-probability -
**Reiny**, Thursday, February 14, 2013 at 7:56amUse the "back-door" approach

number of possible outcomes with no restriction = 9^4 = 6561

number of cases with "no 3" = 8^4 =4096

So the number of cases that have at least some 3's

= 6561-4096 =2465

prob(at least one 3) = 2465/6561 = .3757

or appr 0.38

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