Question

A fair dice is rolled 3 times in a row. The outcomes are shown below.

Calculate the probability of all three events occurring.

Roll Outcome
1st 6
2nd factor of 12
3rd less than 6

Answer 1

25/216

Explanation:

The first roll is obvious. 6 options to a 6 sided dice, one of which is a 6. Since no other option works, 6 is the only correct one, which is a 1/6 chance.

Second roll is a little less obvious. Factors of 12 (in 1-6) are 1, 2, 3, 4, and 6. This is 5 out of the 6 outcomes on a 6 sided dice. This means there is a 5/6 chance to get a factor of 12.

Finally, the roll of the last one can be a 1, 2, 3, 4, or 5, which is 5 out of the 6 outcomes possible on the fair dice. This means there is a 5/6 chance of the outcome being less than 5.

Then multiply the chances together. 1/6, 5/6, 5/6. 1/6 * 5/6 is 5/36. 5/36 * 5/6 is 25/216.

Hope this helps :)

Answer 2

25/216

Explanation:

P( 6) = number of 6's / total number of outcomes

=1/6

Then roll again

factor of 12: 1,2,3,4,6,12 =

1,2,3,4,6 are possible rolls = 5

P ( factor of 12)=number of numbers that are factor of 12 / total number of outcomes = 5/6

Then roll again

P( number less than 6) = numbers less than 6 / total

=5/6

P (6, factors of 12, number less than 6) = 1/6*5/6 *5/6

=25/216