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

You are watching: If a fair 6-sided die is rolled three times, what is the probability that exactly one 3 is rolled?

There room three dice.

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

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

C. Opportunity of getting precisely $~3~$ sixes.$$1/6 * 1/6 * 1/6 = 1/216$$

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



A. There is a total of 6^3=216 combine if you role 3 dice. There space 5^2x3=75 combinations the you will gain one 6. For this reason there is a 75/216=25/72 opportunity of acquiring only one 6 when rolling 3 dice.

B. There room 5x3 combinations the you will acquire 2 6s. Hence there is a 15/216=5/72 possibility of obtaining a 2 6s when rolling 3 dice.

C. There is 1 combination where girlfriend will gain 3 6s. Therefore there is a 1/216 chance you will obtain 3 6s when rolling 3 dice. (Good project you gained this correct)

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


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

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The possibility of getting at the very least one 6 with 3 dice is 91/216 since if friend subtract 125/216 (the probability that rolling three dice without getting a 6) native 216/216 (the probability the any mix of numbers), you acquire 91/216.


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