Probability is defined as a numerical measure which indicates the chance of occurrence. Assume that a coin is tossed the toss may result in the occurrence of head or a tail. The chances of head and tail are equal.
Three systematic approaches are there in probability:
1). Classical approach
2). Empirical approach
3). Axiomatic approach.
The above mentioned approaches has their own merits and demerits.
In the midst of such indefiniteness, predictions are made. This necessitated a systematic study of probabilistic happenings.
Probability theory is considered as a mathematical foundation for statistics and is used in quantitative analysis of large sets of data.
Probabilities are defined to obey certain assumptions and the probability value always lie between 0 and 1.

## The Classical Approach

Elementary probability theory has three different approaches: Classical, empirical and axiomatic. All the different approaches are explained in this page.

Classical definition:
Let a random experiment have mutually exclusive and exhaustive, n equally likely outcomes. Let m of these outcomes be favourable to an event A.
Then the probability of A is
$P(A)$ = $\frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ outcomes}$

=> $P (A)$ = $\frac{m}{n}$

Example 1: What is the probability that 4 cards drawn at random from a well shuffled pack of playing cards belong to different suit?

Solution: A pack has 52 cards. A draw of 4 cards has $^{52}C_{4}$ outcomes. Since the card drawn should be of four different suits, each should be from 13 each cards of different suits.

Therefore $^{13}C_{1}$ * $^{13}C_{1}$ * $^{13}C_{1}$ * $^{13}C_{1}$ outcomes are favourable.

Therefore $P$[4 cards of different suits] =

$\frac{^{13}C_{1} *^{13}C_{1} * ^{13}C_{1} * ^{13}C_{1}}{^{52}C_{4}}$

= $\frac{13*13*13*13}{13*17*25*49}$

= $\frac{28561}{270725}$

= 0.1055

Example 2: Find the probability that a throw of an unbiased die results in
1) An even number.
2) A multiple of 3.

Solution: Sample space $S$ = {1, 2, 3, 4, 5, 6}
There are n = 6 equally likely, exhaustive and mutually exclusive outcomes.
Let events $A$ and $B$ be
$A$ : Throw results in an even number.
$B$ : Throw results in a multiple of 3.

1) Event A has three favourable outcomes, namely, 2, 4 and 6.

$P$ [Even number] = $P(B)$  = $\frac{m}{n}$

= $\frac{3}{6}$

= $\frac{1}{2}$

2) Event $B$ has two favourable outcomes, namely, 3 and 6.

$P$ [Multiple of 3] = $P(C)$ = $\frac{m}{n}$

= $\frac{2}{6}$

= $\frac{1}{2}$

## The Empirical Approach

Let a random experiment be repeated n times essentially under identical conditions. Let m of these repetitions result in the occurrence of an event A.
Then probability of event A is the limiting value of the ratio $\frac{m}{n}$ as n increases indefinitely.

$P(A)$ = $lim$ $_{n\rightarrow \infty}$$\frac{m}{n} Example 1: Out of 785 babies born in a community in a year, 456 were male. Find the probability that a new born baby is male. Solution: Given n = 785 births were observed and among them m = 456 have resulted in birth to male children. Therefore probability that a new born baby is male is P [Male baby] = \frac{m}{n} = \frac{456}{785} = 0.581 ## The Axiomatic Approach Consider a random experiment with sample space S. Let for every event A, a real number P(A) be assigned. Then P(A) is the probability of event A, if the following axioms are satisfied. 1) P(A) \geq 0 2) P(S) = 1, S being the sure event. 3) For two mutually exclusive events A and B, P(A \cup B) = P(A) + P(B) Example 1: If P(A) = 0.8, P(B) = 0.5 and P(A \cup B) = 0.9 find P(\frac{A}{B}). Are A and B independent events? Solution: By addition theorem P(A \cup B) = P(A) + P(B) - P(A \cap B) P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.8 + 0.5 - 0.9 = 0.4 P$$(\frac{A}{B})$ = $\frac{P(A\cap B)}{P(B)}$

= $\frac{0.4}{0.5}$

= 0.8

Thus $P$$(\frac{A}{B}) = 0.8 Here P$$(\frac{A}{B})$= $P(A)$.

Therefore, events A and B are independent.

Example 2: There are 20 persons. 5 of them are graduates. 3 persons are randomly selected from these 20 persons. Find the probability that atleast one of the selected persons is graduate.

Solution: From 20 persons, 3 persons can be selected in $^{20}C_{3}$ ways. Thus, there are $^{20}C_{3}$ possible outcomes. Since there are 15 persons who are not graduates.

$P$ [Atleast one is graduate] = 1 - $P$ [None is graduate]

= 1 - $\frac{^{15}C_{3}}{^{20}C_{3}}$

= 1 - $\frac{91}{228}$

= $\frac{137}{228}$

## Probability Theory Examples

Example 1: A box has 1 red and 3 white balls. Balls are drawn one after one from the box. Find the probability that the two balls drawn would be red if

a) The ball drawn first is returned to the box before the second draw is made. (Draw with replacement).
b) The ball drawn first is not returned before the second draw is made. (Draw without replacement).

Solution :
Let $A$ : First ball drawn is red.
Let $B$ : Second ball drawn is red.

Draw with replacement
$P(A)$ = $\frac{1}{4}$.

Also since the first ball is returned before the second draw is made.

$P$ $(\frac{B}{A})$ = $\frac{1}{4}$

$P$[Two balls are red] = $P(A \cap B)$

= $P(A)$ $\times$ $P$ $(\frac{B}{A})$

= $\frac{1}{4}$ $\times$ $\frac{1}{4}$

= $\frac{1}{16}$

Draw without replacement:

Since the first ball drawn is not returned before the second draw is made.

$P$ $(\frac{B}{A})$ = $\frac{0}{4}$

$P$[Two balls are red] = $P(A \cap B)$

= $P(A)$ $\times$ $P$ $(\frac{A}{B})$

= $\frac{1}{4}$ $\times$ $\frac{0}{4}$

=  0
Example 2: The probability that India wins a cricket match is 0.52. If India plays three matches, find the probability that it wins
1)  Atleast one match
2)  All the three matches.

Solution:
Let $A$ : India wins the first match
$B$ : India wins the second match
$C$ : India wins the third match

Then $P(A)$ = $P(B)$ = $P(C)$ = 0.52

and $P(A')$ = $P(B')$ = $P(C')$ = 0.48  (Because sum of the probabilities is 1)

Here $A$, $B$ and $C$ are independent events.

1) $P$ [Wins atleast one match] = 1 - $P$ [Wins none]

= 1 - $P(A' \cap B' \cap C')$

= 1 - $P(A')$ $\times$ $P(B')$ $\times$ $P(C')$

= 1 - 0.48 $\times$ 0.48 $\times$ 0.48

= 0.8894

2) $P$ [Wins all matches] = $P(A \cap B \cap C)$

= P(A) $\times$ P(B) $\times$ P(C)

= 0.52 $\times$ 0.52 $\times$ 0.52

= 0.1406

Example 3: A card is drawn at random from a pack of cards.
1) What is the probability that it is a heart?
2) If it is known that the card drawn is red, what is the probability that it is a heart?

Solution :
There are total 52 cards in a deck. Let events $A$ and $B$ be
$A$ : Card drawn is red.
$B$ : Card drawn is heart.

There are 26 red cards and 13 hearts in a pack of cards. Therefore, event $A$ has 26 favourable outcomes and event $B$ has 13 favourable outcomes. Event $A \cap B$ has 13 favourable outcomes because when any of the 13 hearts is drawn $A \cap B$ happens.

Therefore, $P(A)$ = $\frac{26}{52}$

$P(B)$ = $\frac{13}{52}$

$P(A \cap B)$ = $\frac{13}{52}$
1) The unconditional probability of drawing a heart is

$P(B)$ = $\frac{13}{52}$

= $\frac{1}{4}$

2) The conditional probability of drawing a heart is

$P$ $(\frac{A}{B})$ = $\frac{P(A \cap B)}{P(A)}$

= $\frac{\frac{13}{52}}{\frac{26}{52}}$

= 0.5

Example 4: A fair coin is tossed thrice. What is the probability of getting 3 heads in each trial.

Solution:
Let events $A$, $B$ and $C$ be
$A$ : The first toss results in head.
$B$ : The second toss results in head.
$C$:  The third toss results in head.

Then $P(A)$ = $P(B)$ = $P(C)$ = $\frac{1}{2}$

Since $A$, $B$ and $C$ are results of three different tosses, they are independent. Therefore, probability that all the three tosses result in head is

$P$[3 heads] = $P(A \cap B \cap C)$

= $P(A)$ $\times$ $P(B)$ $\times$ $P(C)$

= $\frac{1}{2}$ $\times$ $\frac{1}{2}$ $\times$ $\frac{1}{2}$

= $\frac{1}{8}$