Aptitude - Probability - Discussion

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.

If I three throw of dices than what is answer?

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.

How do it?

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

Is there any short method to choose number of event ?

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

Is there any short method to choose number of event ?

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 we got n(e) i.e, 4. Please friends somebody help me to find out the answer.

@Rama.

So n(e)=4. It's because the possible combinations where you would get sum 9 when you throw the dice twice are 4.

Like in a dice there are 1-6 no.s right? So what no.s do you put together to get sum 9? (3,6), (4,5), (6,3) ,(5,4) .

But these combos aren't possible: (2,7), (1,8) because well they're not on a dice.

Hope THAT cleared it up for you! ^_^.