When we do something we call it an experiment and the outcomes of that event are taken as separate events. For example: throwing a dice is an experiment and getting a multiple of 2 as outcome is one of the possible events. The ratio of total number of favorable outcomes to an event to the total number of outcomes possible in that experiment is known is probability of that event. In simple words probability is just the possible chances of occurrence of an event. Probability always lies between 0 and 1. When probability is equal to zero, then the event is said to be an impossible event. When the probability is equal to one, we call the event a sure event.

When we are calculating probability of any event using the formula as stated above that is probability of an event is the ratio of number of favorable outcomes of the event to that of the total outcomes of that experiment, we are actually finding theoretical probability. This is because we are just making assumptions and finding the probability but actual probability may vary in practical case. 
While experimental probability is calculated on the basis of repeated performance of the experiment to large number of times to attain more accuracy and precised probability, the theoretical probability is simply calculated on the basis of fair possibilities. Experimental probability is based on practical performance, observations and experiences.

In that case, the experimental probability also known as empirical probability is evaluated as the ratio of the number of times the event has occurred to the total number of times the occurrences are observed.
Let E be an experiment that is performed and A be an event associated to it. 

Let n (E) be the total number of outcomes of the experiment and n (A) be the number of outcomes favorable to the event A.

Then the probability of occurrence of event A is denoted by P (A) is given by:

$P\ (A)$ = $\frac{[n\ (A)]}{[n\ (E)]}$
When we represent the probabilities of all possibilities of any random variable we call it a probability distribution. In this case, when we make use of theoretical probabilities too make the distribution we call it theoretical probability distribution. Binomial distribution, Poisson distributions of probability are examples of theoretical probability distribution. In a Binomial distribution, we assume a fix number of trials ‘n’ being carried out in which only two outcomes can be possible: either a success or a failure. Also, here the probability of failure is equal to probability of success subtracted from 1. The trials in this case have to be independent, that is, the outcome of one trial should not affect the outcome of the next trial at all. Then we have a Binomial distribution on random variable X, as Bin (n, p), where ‘n is the number of trials of ‘p’ is the probability of success.
The formula for Binomial distribution is given by:

$P\ (X\ =\ x)\ =\ (n,\ x)\ p^{x}\ (1\ –\ p)^{(n\ –\ x)}$

Where, $(n,\ x)$ = $\frac{n!}{(x!\ (n\ –\ x)!)}$

Here, $n!\ =\ 1\ .\ 2\ .\ 3\ …\ (n\ –\ 1)\ .\ n$
In case of Poisson distribution, if X is representing the number of events in a given interval, then we have that the single event is occurring within that interval is actually proportional to the length of that interval. Also, the even can occur infinite number of times is also a possibility. Then by Poisson distribution, 
$P\ (X\ =\ x)$ = $\frac{[e^{(-\lambda)}\ .\ \lambda^{x}]}{(x!)}$
Where, $\lambda$ is the number of occurrences that are expected for the event in the given interval.
Also, $\lambda$ = $n\ .\ p$
In particularly large values of n’ we make use of Poisson distribution.
Let us see an example based on theoretical probability.
Example 1:

In an experiment two dice are thrown. Let A be the event of getting same number on both dice, B be the experiment of getting a total of 7 and C be the event of getting odd numbers on both dices. Find the probabilities of events A, B and C.

Solution:

In this case total number of possible outcomes $n\ (E)\ =\ 6\ \times\ 6\ =\ 36$

Favorable outcomes for event $A\ =\ {(1,\ 1),\ (2,\ 2),\ (3,\ 3),\ (4,\ 4),\ (5,\ 5),\ (6,\ 6)}$

$\Rightarrow\ n\ (A)\ =\ 6$

Favorable outcomes for event $B\ =\ {(1,\ 6),\ (2,\ 5),\ (3,\ 4),\ (4,\ 3),\ (5,\ 2),\ (6,\ 1)}$

$\Rightarrow\ n\ (B)\ =\ 6$

Favorable outcomes for event $C\ =\ {(1,\ 3),\ (1,\ 5),\ (1,\ 1),\ (3,\ 1),\ (3,\ 3),\ (3,\ 5),\ (5,\ 1),\ (5,\ 5),\ (5,\ 3)}$

$\Rightarrow\ n\ (C)\ =\ 9$

$P\ (A)$ = $\frac{6}{36}$ = $\frac{1}{6}$

$P\ (B)$ = $\frac{6}{36}$ = $\frac{1}{6}$

$P\ (C)$ = $\frac{9}{36}$ = $\frac{1}{4}$