Two fair dice are thrown.How do you find the probability that the sum of the numbers thrown is not less than 6 ?

I'm not sure how else to do this question except by writing it all out.
Because the number we are summing to is small, there's less ways they can add to less than 6, than it not happening. Therefore its easier to calculate the probability of this not happening and then subtract that from 1.
ie. We want to calculate the probability that the sum of 2 die is less than 6. Which is the same as adds to 5 or less (because we dont want the ones equal to 6).

My next step would be to figure out how many different ways the dice could fall. This is equal to the number of ways the first dice could fall multiplied by the number of ways the second dice could fall. If you have trouble picturing this sometimes a tree diagram helps.
ie. #6*6=36#

Its important to remember that if the first dice rolls a 1 and the second dice rolls a 3 this is a different event to if the second dice rolls a 1 and the first dice rolls a 3.
This means that for every combination of numbers we come up with it could happen in the opposite way as well, except if the same number comes up twice
eg one dice shows a 2 and the other also shows a 2
((you can skip this by writing out both outcomes separately but these shortcuts will make life easier with longer and more difficult questions))

Then I would systematically list the outcomes I'm after, making sure I don't miss any.
Outcomes that can happen in 2 ways : 1&2, 1&3, 1&4, 2&3. Total=4
Outcomes that can happen 1 way : 1&1, 2&2. Total=2

Then I would multiply the events which can happen 2 ways by 2 (because they are twice as likely to happen) and add the numbers of outcomes together.
#4*2+2=10#

Then the probability is the number of desired outcomes over the number of total possible outcomes
#10/36=5/18#

Remember though that this is still the probability of the complementary event (the opposite of what we want).
To find the probability we then subtract this from 1
#1-5/18=13/18#