Aptitude - Probability - Discussion

In that question they have asked for 2 dices, it means each dice is having 6 sides. So for 2 dices= 6*6=36. n(S)=36

Then we should find the possibility for sum of 9 so we can take (4,5),(3,6),(6,3),(5,4). These are 4 possibilities for sum of 9. n(E)=4.

After that Probability of E = n(E)/n(S) = 4/36 = 1/9.

How do it?
In throws of a dice, n(s) = (6*6) = 36.

Is there any short method to choose number of event ?

Is there any short method to choose number of event ?

How do it?

In two throws of a dice, n(S) = (6 x 6) = 36.

I understood of this way I don't know this is right or wrong.

We want to getting some 36 = 36+9 = 45.
45+9 = 54.
54+9 = 63.

If I three throw of dices than what is answer?

I have a little bit confusion. Question is "What is the probability of getting a sum 9 from two throws of a dice?". Here we talk about two throws means that a dice is thrown 2 times.

Further it means that a single dice is throw 2 times. So, we should take sample space of 6 because of a single dice.

In case of 36 the question should be "What is the probability of getting a sum 9 when two dice are thrown?". Or perhaps I am in mistake in understanding the question. Please clear this.

Is there any short method to choose number of event ?

1,1 2,1 3,1 4,1 5,1 6,1
1,2 2,2 3,2 4,2 5,2 6,2
1,3 2,3 3,3 4,3 5,3 6,3
1,4 2,4 3,4 4,4 5,4 6,4
1,5 2,5 3,5 4,5 5,5 6,5
1,6 2,6 3,6 4,6 5,6 6,6

P(sum 9) = {(6,3),(5,4),(4,5),(3,6)}.

= 4/36.

= 1/9.