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October 25, 2014

October 25, 2014

Posted by **Amanda** on Wednesday, May 2, 2012 at 5:05pm.

- Probability -
**MathMate**, Wednesday, May 2, 2012 at 5:34pmThe two dice are assumed to give random and independent results.

So the outcomes for each die is equally probable.

You can construct a 6x6 table containing all 36 possible outcomes, and pick out those with a sum of 3 or 5. The total number of these favourable outcomes divided by 36 gives the required probability.

You are probably familiar with combinatoric counts of the outcomes totaling 3 or 5. The count divided by 36 also gives the required probability.

- Probability -
**HUTI**, Friday, October 24, 2014 at 11:32am!@#$%^& U

- Probability -
**HANDO**, Friday, October 24, 2014 at 11:33am!@#$%^&

- Probability -
**HANUR**, Friday, October 24, 2014 at 11:33amPEPEK KAu

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