If two balanced dice are thrown simultaneously, write the sample space and also find the probability that the sum of two numbers appearing on the top of the dice is less than or equal to 12.

Hint: Here to solve the given question we need to first write the sample space, we know that sample space has the elements which are possible for the given event, as here two dice are rolled so accordingly we have to make the sample space. To find the probability for the event we need to find the ratio of favorable events to the total possible space.
Formulae Used: Probability for the event is:
\[ \Rightarrow \dfrac{{favourable\,terms}}{{total\,terms}}\]

Complete step-by-step solution:
Here the given question is to find the sample space, and then find the probability for the event, on solving we get:
Sample space:
\[
   \Rightarrow \{ (1,1)(1,2)(1,3)(1,4)(1,5)(1,6)(2,1)(2,2)(2,3)(2,4)(2,5) \\
  (2,6)(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)(4,1)(4,2)(4,3)(4,4)(4,5) \\
  (4,6)(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)(6,1)(6,2)(6,3)(6,4)(6,5) \\
  (6,6)\} \\
 \]
Here we get 36, possibilities in the sample space, when two dice are rolled.
Now the favorable events, have the condition that sum of the element are less then or equal to twelve:
Hence the favorable events are, the every events of the sample space because the maximum sum of the events is, twelve and every event sum is either equal to or less then twelve which is the condition given in the question, hence probability would be:
\[ \Rightarrow P = \dfrac{{36}}{{36}} = 1\]
Here the probability is one.

Note: Here the given question is to solve for the probability for the given event, and to find this we need to find the possible events, and the total possible events, the ratio for both the terms would give the probability for an event.