What is the probability that two die when tossed will end up with the sum of $8$ on top?

Hint: Every dice has six sides which are written as $1,2,3,4,5,6$ and when we toss two die then the total number of possible ways that can come up is $36$ and the sum of $8$ can come up in the following ways which are $(2,6),(3,5),(4,4),(5,3),(6,2)$ and hence we can find the probability.

Complete step-by-step answer:
Here we need to find the probability that two die when tossed will end up with the sum of $8$ on the top. So basically dice is the small cube which has a different number on the different six sides of the cube which are numbered from $1{\text{ to 6}}$ and here the minimum number is $1$ and the maximum number that can come is$6$. So basically if we are throwing two dice then at the same time we may get $(1,1)$ which means that $1$ on the first and $1$ on the second dice too. So we get the total of two as a sum when both the dice are tossed.
But we want that the sum on the numbers on both the die must be equal to $8$
First off all we must know all the $36$ cases that may be possible if we throw the two die which will be
$
  \{ (1,1)(1,2)(1,3)(1,4)(1,5)(1,6) \\
  (2,1)(2,2)(2,3)(2,4)(2,5)(2,6) \\
  (3,1)(3,2)(3,3)(3,4)(3,5)(3,6) \\
  (4,1)(4,2)(4,3)(4,4)(4,5)(4,6) \\
  (5,1)(5,2)(5,3)(5,4)(5,5)(5,6) \\
  (6,1)(6,2)(6,3)(6,4)(6,5)(6,6)\} \\
 $
So in this we have all the cases that are possible outcomes that can come in the throw of two die. But we want only those cases where the sum is equal to $8$
So form the above cases we come to know that the sum of the die on both the outcomes is $8$ in only the following cases which are $(2,6),(3,5),(4,4),(5,3),(6,2)$
So there are only $5$ above cases where the sum comes out to be $8$ on the die.
So the probability is the ratio of the number of required outcomes to the total number of possible outcomes that can come so required probability is $\dfrac{5}{{36}}$

Note: Similarly if we are given that the three coins are tossed and we need to find any case then the possible outcomes will be $\{ HHH,HTH,HTT,THT,TTH,HHT,THH,TTT\} $ and here T represents the tail on the coin and H represents that head appears on the coin and there are $8$ possible outcomes.