Aptitude - Probability - Discussion

Ankita:.

Because 6c1 * 6c1 is 36.

s is a sample space i.e all possible outcomes
if two coins are tossed then s={HH,TT,HT,TH}
same in this qu. two dice -use then no of possible outcome
s={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)},
{(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)}
..........
{(6,1),(6,2),(6,3),(6,4),(6,5),(6,5)}

n(s)=6+6+6+6+6+6=36
or
6*6=36

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.

Is there any short method to choose number of event ?

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 ?

Is there any short method to choose number of event ?