ns am having actually a hard time pack my head about this and am certain that mine answers are wrong.

You are watching: Probability of rolling a 6 with 3 dice

There room three dice.

A. Possibility of getting precisely one six on the three dice.$$(1/6) * 3 = 1/3$$

B. Opportunity of getting specifically two sixes.$$(1/6 * 1/6) * 1.5 = 1/24$$

C. Chance of getting specifically $~3~$ sixes.$$1/6 * 1/6 * 1/6 = 1/216$$

D. Chance of any combination of A, B and also C$$1/3 + 1/24 + 1/216=72/216 + 9/216 + 1/216 = 82/216$$



A. There is a total of 6^3=216 combinations if you role 3 dice. There space 5^2x3=75 combinations that you will obtain one 6. Therefore there is a 75/216=25/72 possibility of getting only one 6 once rolling 3 dice.

B. There space 5x3 combinations the you will acquire 2 6s. Therefore there is a 15/216=5/72 chance of getting a 2 6s as soon as rolling 3 dice.

C. Over there is 1 mix where friend will acquire 3 6s. Hence there is a 1/216 chance you will acquire 3 6s when rolling 3 dice. (Good task you got this correct)

D. Over there is a 75+15+1/216=91/216 possibility of any type of of lock happening.


Hint because that A: what is the possibility that no sixes appear? If you know that opportunity then you automatically know the chance that sixes do appear.

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The possibility of getting at least one 6 with 3 dice is 91/216 due to the fact that if friend subtract 125/216 (the probability the rolling 3 dice without acquiring a 6) native 216/216 (the probability of any mix of numbers), you gain 91/216.


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