The face of a 12-sided die are numbered from 1 to 12. a) What is the probability that the first ten will be on the third roll?    explain please.

Asked on by zofic13

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lfryerda | High School Teacher | (Level 2) Educator

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If you have more than one question you need to make separate posts.

Each roll has 12 possibilities, so there are `12 times 12 times 12 = 1728` different number of rolls.

Since the first 10 must be on the third roll, there are 11 possibilities for each of the first and second rolls, with 1 possibility (a 10) on the third roll.  This gives `11 times 11 times 1= 121` different rolls.  This means the probability is given by 

`121/1728 approx 0.07002 = 7.002%`

The probability is `121/1728` .

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