We roll two fair dice three times.
(i) What is the probability that we get consecutive sums $6, 8, 4?$
(ii) What is the probability that we get a sum of $4$ on the third throw if on the first two we get a sum of $8$ and $5$ respectively?
(iii) What is the probability that the three outcomes contain exactly one sum of $4$ and exactly one sum of $10?$
My answer to (i) was: The total probability of getting these sums is: $\frac{5}{36}\cdot\frac{5}{36}\cdot \frac{3}{36}\approx 0.0016$. But I'm not exactly sure how to find the answer to the other two. Does (ii) imply that the throws are somehow conditionally bound? Wouldn't it normally be the probability of getting a sum of $4$ irregardless of what you have thrown before? (so $\frac{3}{36}$). And I'm not sure how to approach the third question aswell. Thanks, in advance.
That is something you have to determine. And you did that.
Indeed. Each pair of rolls are independent of previous rolls.
You have the probability for a particular roll being 4, 10, or elsewise. Being $3/36$, $3/36$, and $30/36$ respectively.
You should be able to count the arrangements of these three distinct results. $\{[4,10,E], [4,E, 10],\ldots, [E,10,4]\}$
Go forth and multiply.