A fair dice is throw twice find the probability that the sum obtain will be a factor of 12?

To find the probability that the sum obtained will be a factor of 12, we first need to determine all the possible outcomes when throwing a fair dice twice.

There are 6 possible outcomes for each throw of the dice, so there are a total of 6 x 6 = 36 possible outcomes when throwing the dice twice.

Next, we need to determine the factors of 12. The factors of 12 are 1, 2, 3, 4, 6, and 12.

Now, we need to find all the combinations of outcomes that will result in a sum that is a factor of 12:

- Sum of 1: (1, 1)
- Sum of 2: (1, 1)
- Sum of 3: (1, 2), (2, 1)
- Sum of 4: (1, 3), (2, 2), (3, 1)
- Sum of 6: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1)
- Sum of 12: (6, 6)

There are a total of 11 combinations that result in a sum that is a factor of 12.

Therefore, the probability of obtaining a sum that is a factor of 12 when throwing a fair dice twice is 11/36 or approximately 0.3056.

Total number of outcomes when a fair dice is thrown twice $$= 6\times 6 = 36$$ Favourable outcomes when the sum is a factor of 12 are $$ 2+10, 3+9 $$ that is 4 Therefore, the probability $$= \frac {Number\quad of\quad favourable\quad outcomes}{Total\quad number\quad of\quad outcomes}=\frac{4}{36}=\frac{1}{9}$$