# Definition of Probability

Probability is unusual and it has been defined in different manners. We can define probability in objective or subjective terms. Let us first use the objective approach to define probability.

The Classical Definition of Probability

This definition is for equally likely outcomes. If an experiment can produce $N$ mutually exclusive and equally likely outcomes, out of which $n$ outcomes are favorable to the occurrence of event $A$, then the probability of $A$ is denoted by $P\left( A \right)$ and is defined as the ratio $\frac{n}{N}$. Thus the probability of $A$ is given by
$P\left( A \right) = \frac{{{\text{Number of outcomes favorable to }}A}}{{{\text{Number of possible outcomes}}}} = \frac{n}{N}$

This definition can be applied in a situation in which all possible outcomes and the outcomes in the events $A$ can be counted. This definition is due to P.S. Laplace (1749 – 1827). The classical definition is also called the priori definition of probability. The word priori is from prior, and is used because the definition is based on the previous knowledge that the outcomes are equally likely. When a coin is tossed, the probability of heads is assumed to be $\frac{1}{2}$. This probability of $\frac{1}{2}$ is based on this classical definition of probability. It is assumed that the two faces of the coin are equally likely. In practical terms, people do believe that a coin will do justice when it is tossed. On the playground, the participating teams toss the coin to start the match. A coin in which probability of heads is assumed to be equal to the probability of tails is called a true or uniform or an unbiased coin. But it is an all an assumption. The probability of a certain event is a number which lies between $0$ and $1$. If the event does not contain any outcome, it is called an impossible event and its probability is zero.

If the event is as big as the sample space, the probability of the event is one because
$P\left( {{\text{the event}}} \right) = \frac{{{\text{Number of outcomes in the event}}}}{{{\text{total number of outcomes}}}} = \frac{N}{N} = 1$

When the probability of an event is one, it is called a “sure” or “certain” event.

Criticism
The classical definition of probability has always been criticized for the following reasons:

1. This definition assumes that the outcomes are equally likely. The term equally likely is almost as difficult to define as the word probability itself. Thus the definition uses circular reasoning.
2. The definition is not applicable when the number of outcomes is not equally likely.
3. The definition is also not applicable when the total number of outcomes is infinite or it is difficult to count the total outcomes or the outcomes favorable to the event. It is difficult to count the fish in the ocean. Thus it is difficult to find the probability of catching a fish of some weight, say more than one kilogram.

Example:

One day 20 files were presented to an income tax officer for disposal. Five files contained false entries. All the files were thoroughly mixed and there was no indication on the false files. What is the probability that one file with false entries is selected?

Solution:
Here all possible outcomes $= 20$

Let $A$ be the event that the file has false entries.

Thus, the number of favorable outcomes $= 5$

Here we shall apply the classical definition of probability. All $20$ files are assumed to be equally likely for the purpose of selecting a file.

The probability of selecting a file with false entries is written as $P\left( A \right)$
$P\left( A \right) = \frac{{{\text{Number of outcomes favorable to }}A}}{{{\text{Number of possible outcomes in }}S}} = \frac{{n\left( A \right)}}{{n\left( S \right)}}$
$= \frac{5}{{10}} = \frac{1}{4}$

Example:

A fair die is thrown. Find the probability that the face on the die is (1) Maximum (2) Prime (3) A multiple of $3$ (4) A multiple of $7$

Solution:

There are $6$ possible outcomes when a die is tossed. We assume that all $6$ faces are equally likely. The classical definition of probability is to be applied here.

The sample space is $S = \left\{ {1,2,3,4,5,6} \right\}$,    $n\left( S \right) = 6$

(1) Let $A$ be the event that the face is maximum
Thus,
$A = \left\{ 6 \right\}$,          $n\left( A \right) = 1$

Therefore, $P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( S \right)}} = \frac{1}{6}$

(2) Let $B$ be the event that the face is prime
Thus,
$B = \left\{ {2,3,5} \right\}$,           $n\left( B \right) = 3$

Therefore, $P\left( B \right) = \frac{{n\left( B \right)}}{{n\left( S \right)}} = \frac{3}{6} = \frac{1}{2}$

(3) Let $C$ be the event that the face is a multiple of $3$
Thus,
$C = \left\{ {3,6} \right\}$,  $n\left( C \right) = 2$

Therefore, $P\left( C \right) = \frac{{n\left( C \right)}}{{n\left( S \right)}} = \frac{2}{6} = \frac{1}{3}$

(4) Let $D$ be the event that the face is a multiple of $7$
Thus,
$D = \phi$,       $n\left( D \right) = 0$

Therefore, $P\left( D \right) = \frac{{n\left( D \right)}}{{n\left( S \right)}} = \frac{0}{6} = 0$   (not possible)