empirical probability P(E) of an event E as
P(E)= Number of trials in which the event happened / Total number of trials

The theoretical probability (also called classical probability) of an event E, written as P(E),
P(E)= Number of outcomes favourable to E / Number of all possible outcomes of the experiment

Probability of throwing of dice

A dice has six (6) faces with number 1,2,3,4,5,6 on it.
So, on throwing of dice, we can get either 1,2,3,4,5 or 6 .
Therefore number of all possible outcome is 6.

SO, the probability of getting any number will be $1/6$
SO, the probability of getting any two number will be $2/6$
SO, the probability of getting any three number will be $3/6$
SO, the probability of getting any four number will be $4/6$
SO, the probability of getting any five number will be $5/6$
SO, the probability of getting any six number will be $6/6$

Probability of getting number less than or equal to 1 => $1/6$
Probability of getting number less than or equal to 2 => $2/6$
Probability of getting number less than or equal to 3 => $3/6$
Probability of getting number less than or equal to 4 => $4/6$
Probability of getting number less than or equal to 5 => $5/6$
Probability of getting number less than or equal to 6 => $6/6$

Probability of getting even numbers = > $3/6$ => $1/2$
Probability of getting odd numbers = > $3/6$ => $1/2$

Probability of throwing of two dice

If two dice is thrown simantenously , the toal number of possible outcome will be,
=> possible outcome of 1st dice is 6 
=> and possible outcome of 2nd dice is 6.
Therefore possible outcome of two disc will be  6x6 = 36.

What is the probability of getting the same number(doublets) in the two dices 
Solution 
Possible outcome of the event of getting the same number on two dices,
i.e (1,1) or (2,2) or (3,3) or (4,4) or (5,5) or (6,6) is 6.

P(E)= Number of outcomes favourable to E / Number of all possible outcomes of the experiment 
P(E) = $6/36$ = $1/6$

What is the probability of getting different numbers in the two dices?
Solution:
Probability of getting different numbers in the two dices 
        = 1- probability of getting same numbers  in the two dices.
        = $1- 6/36$		
        = $30/36$
        = $5/6$