Two dice have been thown, giving a total of at least 10. What is the probability that the throw of a third die

Two dice have been thown, giving a total of at least 10. What is the probability that the throw of a third die will bring the total of the three numbers shown to 15 or higher?

This isn't a simple question. The point here is that the first two dice didn't score 10 they scored at least 10.
so they scored 10, 11 or 12
Now we have to work each probability out
There are 3 ways of scoring 10 (5 5, 4 6 or 6 4)
there are 2 ways of scoring 11 (6 5 or 5 6)
There is one way of scoring 12 (6 6)

sO P(10) = 3/6
P(11) = 2/6
p(12) = 1/6

Now if he scores 10 he needs to get a 5 or a 6 which is 2/6 so P(15+) = 3/6*2/6 = 6/36
Now if he scores 11 he needs to get a 4,5 or 6 which is 3/6 so P(15+) = 2/6 * 3/6 = 6/36
Now if he scores 12 he needs to gat a 3,4,5 or 6 which is 4/6 so P(15+) = 1/6* 4/6 = 4/36

total probability is 6/36 |+ 6/36 + 4/36 = 16/36 = 4/9

For every die thrown, there is a 1/6 probability for each face, regardless of previous throws. So, since we are only focusing on the third die, we know that there is a 1/6 chance of rolling a 5, and a 1/6 chance of rolling a 6. Therefore there is a 2/6 chance (or 1/3 chance) of rolling a 15 or higher.