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## Homework Statement

Explain the formulas used to obtain the solution for the question above (What is the probability of two

**tetrahedral**dice landing on a 6 if twelve dice are rolled?)

1 - (3/4)^12 - 12 * ((3/4)^11) * (1/4) = 0.842

## Homework Equations

If we were looking for one dice landing on 6 (out of 12 dice) it would be

1 - (3/4)^12

## The Attempt at a Solution

So, I think that I understand that (3/4)^12 is 12 dice that are NOT 6 (unsuccessful outcomes) and that a dice landing on a six would be 1/4 (a successful outcome). This is because a tetrahedral die has 4 sides (3 out of 4 sides would be unsuccessful, while the 1 out of 4 would be successful/ a 6

Further in the formula, (3/4)^11 is 11 out of 12 dice that don't land on 6 and 1/4 is that remaining dice that land on a 6. This is multiplied by 12 because there are 12 dice and any one can land on 6 (while the other 11 are a non-6).

Why is 12 * ((3/4)^11) * (1/4) being subtracted from

1 - (3/4)^12?[/B]

I hope I'm explaining this right! Thank you

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