Probability that sum must be at least 6 $ = 1 – P(sum < 6)$
Case1:
4, 5 are the only terms that are less than 6, and each has a probability of $1/6$, and hence $1/3$
Case 2:
Now, when the value on die is $1, 2, 3 $, the total favourable outcomes are:
$(1,1), (1, 2), (1, 3), (1, 4)$
$ (2, 1), (2, 2), (2, 3)$
$(3, 1), (3, 2),$
out of total 36 outcomes.
So, $P(sum < 6)$ = $P(case 1) + P(case 2)$
$ = 7/12$
$P(sum >= 6) = 1 – P(sum < 6)$
$= 1 – 7/12$
$ = 5/12$