**15.1. INTRODUCTION** In our everyday life we sometimes make statements for the possible future events by using words ‘probably’, ‘doubt’, ‘chances’, ‘most probably’, ‘likely’, etc. Using these words we predict for any future event based on our past experiences under similar conditions i.e. we try to measure degree of uncertainty. Probability is the science, which measure degree of uncertainty numerically.

Though probability started with gambling, it has been used extensively in many fields like weather forecasting, commerce, biological sciences, medical sciences, physical sciences, etc.

**Remarks**

The measure of uncertainty is probability.

**History of Probability**

The theory of probability was originated from the games of chance related to gambling. An Italian Mathematician, Jerome Cardan (1501–1576) was the first to write a book named “Book on Games of Chance” published in 1663. Notable contributions were also made by mathematicians J. Bernoulli, P. Laplace and A. A. Markov. In the twentieth century, a book “Foundation of Probability” was published by Russian Mathematician Kolomogorov in 1933 and this was the first book to introduce probability as a set function.

**Some Important Objects**

(i) Coin : Coin is a well known object. It has two faces, one is Head and other is Tail.

(ii) Die : A die is a well balanced solid cube having six faces marked with numbers (dots) from 1 to 6, one number of one face. The plural of die is dice.

(iii) Playing Cards : A pack of playing cards contains 52 cards out of which 26 are red cards and 26 are black cards.

These 52 cards are divided in four groups, each group is called a suit and has 13 cards. Name of these suits are:

(i) Diamond ($\u25c6$)

(ii) Heart ($\u2665$)

(ii) Spades ($\u2660$)

(iv) Club ($\u2663$)

Out of these four suits Diamond and Heart are read cards and Spade and Club are black cards. Each suit having 13 cards which are 1, 2, 3, …., 10, Jack, Queen and King. Card having 1 is also called an ace. Jack, Queen and King are known as face cards. Therefore total 12 face cards are in a pack of 52 playing cards.

Some definitions :

**15.2. EXPERIMENT**

An activity which ends in some well defined results is called an experiment. These results are called outcomes. There are two types of experiments :**(i) Deterministic experiment**Those experiments which when repeated under identical conditions produce the same results or outcomes are known as deterministic experiments.

Example: Formation of Methane in laboratory.

**(ii) Random Experiment**

An experiment, when repeated under identical conditions do not produce the same outcome every time but the outcome is one of the several outcomes, it is known as Random Experiment:

Example :

(i) throwing a die

(ii) tossing a coin

(iii) drawing a card from a pack of playing cards

(iv) selecting a student in a class without preference.

Probability deals with Random Experiments.

**Trial**

Performing an experiment once is called a trial.

**Sample Space**

The collection of all the possible outcomes of a random experiment is called a sample space. It is usually denoted by S.

Example : After tossing a coin, possible outcomes are head and tail so sample space for tossing a coin consists of head and tail.

**Event**

Each possible outcome of a trial is known as an event. It is generally denoted by E. It is of two types:

**(i) Simple Event :**If any event E contains only one outcome of sample space then it is known as simple event. In this way each outcome of sample space related to any experiment is a simple event.

**Example :**The experiment of throwing a die once consists of 6 simple events viz. coming to the face showing up 1 or 2 or 3 or 4 or 5 or 6.

**(ii) Compound Event :**If any event contains more than one outcomes of sample space, then it is known as compound event.

**Example :**After throwing a die the outcome is an even number i.e. 2 or 4 or 6.

**15.3. DIFFERENT APPROACHES TO PROBABILITY**

There are following approaches to theory of probability :

(1) Empirical Approach

(2) Classical Approach

(3) Axiomatic Approach

Here we study only Empirical Approach to Probability**Definition of Probability**Let

*E*be any event related to a Random experiment whose sample space has

*n*outcomes and out of these

*n*outcomes, the event can be performed by

*m*outcomes, then probability of occurrence of event

*E*will be

P(E)$=\frac{NumberoffavourableoutcomesinsamplespaceforeventE}{Totalnumberofoutcomesinsmplespace}$

$P\left(E\right)=\frac{m}{n};0\le m\le n$

$Itsclear,\mathrm{sin}ce0\le m\le n$

$\Rightarrow 0\le \frac{m}{n}\le 1$

$\Rightarrow 0\le P\left(E\right)\le 1$

i.e. probability of any event lies between 0 and 1.**For Example :** If we toss a coin then there are two possible outcomes in sample space one is Head (H) and other is Tail (T).

So we have n = 2, out of these two outcomes one outcome is in favour of event ‘coming up head’ and one outcome is in favour of event ‘coming up tail’. Therefore, probability of getting a head.

P(H) = $\frac{1}{2}$ [ $\therefore $ m = 1]

and, probability of getting a tail,

P(T) = $\frac{1}{2}$ [ $\therefore $ m = 1]

Similarly, when we throw a die then number of possible outcomes = 6

Therefore, *n* = 6

Now if E is the event of coming an even number then it can be happened by coming of 2 or 4

or 6.

Therefore, number of favourable cases = *m* = 3

So, probability of getting an even number = $\frac{m}{n}=\frac{3}{6}=\frac{1}{2}$*Note :* If *n* be the total number of trials and *m* be the number of trails in which the event E can be happened then probability of occurrence of event E will be

OP=$\sqrt{{x}^{2}+{y}^{2}}=\frac{m}{n}$

For example, if a coin is tossed for 250 times and 170 times we got head then probability of getting a head will be

P(H) = $\frac{m}{n}$

= $\frac{170}{250}$ [ $\therefore $ m = 170 and n = 250] = $\frac{17}{25}$ .** Remarks**(i) Probability of any event cannot be less than 0 and cannot be more than 1. So it can be any fraction from 0 to 1.

(ii) It p is the probability of occurrence of an event E and q is the probability of non occurrence of that event then

p + q = 1

$\Rightarrow $q = 1 – p

(iii) The sum of the probabilities of all the possible outcomes of a trial is 1.

Impossible Event

If the number of favourable outcomes for an event is zero then the probability of occurrence of that event will be zero and such type of event is known as Impossible Event.

**15.4. SURE EVENT OR CERTAIN EVENT**

It the number of favourable outcomes for an event is equal to the total number of possible outcomes then the probability of occurrence of that event will be one and such type of event is known as sure or certain event.**Example 1**1500 families with 2 children were selected randomly, and the following data were recorded:

Number of girls in a family |
2 |
1 |
0 |

Number of families |
475 |
814 |
211 |

Compute the probability of a family, chosen at random, having

(i) 2 girls

(ii) 1 girls

(iii) No girls

Also show that sum of all these probabilities is 1.**Solution**Total number of families = n = 1500

(i) Number of families having two girls = m = 475

Let ${E}_{1}$ be the event that the family, chosen at random, is having 2 girls, then

P (${E}_{1}$) =$\frac{Numberoffamilieshavingtwogirls}{Totalnumberoffamilies}$

$\frac{m}{n}=\frac{475}{1500}=\frac{19}{60}$

Hence P(${E}_{1}$) =$\frac{19}{60}$

(ii) Number of families having one girl = m = 814

Let E_{2} be the event that the family, chosen at random, is having one girl, then

P(E_{2}) =$\frac{Numberoffamilieshavingonegirl}{Totalnumberoffamilies}$

$\frac{m}{n}=\frac{814}{1500}=\frac{407}{750}$

Hence, P(${E}_{2}$) =$\frac{407}{750}$

(iii) Number of families having no girls = m = 211

Let E_{3} be the event that the family, chosen at random, is having no girl, then,

P(E_{3}) =undefined

$\frac{m}{n}=\frac{211}{1500}$

i.e. P(${E}_{3}$) =$\frac{211}{1500}$

Now, P(E_{1}) + P(E_{2}) + P(E_{3})

$=\frac{19}{60}+\frac{407}{750}+\frac{211}{1500}\phantom{\rule{0ex}{0ex}}=\frac{475}{1500}+\frac{814}{1500}+\frac{211}{1500}=\frac{475+814+211}{1500}=\frac{1500}{1500}=1$

Hence sum of all these probabilities is 1.

**Example 2**A bag contains 20 cards numbered from 1 to 20. One card is drawn from the bag. Find the probability that it bears a prime number.

**Solution**

Here, total number of possible outcomes = n = 20

Prime numbers between 1 to 20 are 2, 3, 5, 7, 11, 13, 17, 19

So, number of favourable outcomes = m = 8

If E is the event of getting a prime number card when a card is drawn from the bag containing 20 cards numbered from 1 to 20, then

P(E) $=\frac{m}{n}=\frac{8}{20}=\frac{2}{5}$

$\therefore $ Required probability = $\frac{2}{5}$ .

**Example 3**A box contains 50 balls of the same shape and weight. Among the balls 10 balls are white, 15 are red and rest are black. A ball is randomly drawn from the bag. Find the probability that the ball drawn is

(i) White

(ii) Black or Red

(iii) Not black**Solution**Total number of balls in box = n = 50

(i) Number of white balls = m = 10

If ${E}_{1}$ is the event of drawing a white ball, then

$P\left({E}_{1}\right)=\frac{m}{n}=\frac{10}{50}=\frac{1}{5}\phantom{\rule{0ex}{0ex}}i.e.P\left({E}_{1}\right)=\frac{1}{5}$

Hence, probability of drawing a white ball = 1/5.

(ii) Number of red balls in bag = 15

and number of black balls = 50 – (10 + 15) = 25

So, favourable outcomes for getting a red or black ball

= m = 15 + 25 = 40

If ${E}_{2}$ is the event of drawing a black or red ball,

Then, P(${E}_{2}$) = $\frac{m}{n}=\frac{40}{50}=\frac{4}{5}$

i.e. P(${E}_{2}$) =$\frac{4}{5}$

Hence, probability of drawing a black or red ball =$\frac{4}{5}$

(iii) Number of balls, which are not black = m = (10 + 15) = 25

If ${E}_{3}$ is the event of drawing a ball which is not black

Then, P(${E}_{3}$) = $\frac{m}{n}=\frac{25}{50}=\frac{1}{2}$

i.e. P(${E}_{3}$) = 1/2

Hence, probability of drawing a ball which is not black = $\frac{1}{2}$.

**Example 4**Three coins are tossed at a time. Find the probability of getting

(i) 2 heads

(ii) atleast 2 heads

(iii) atmost 2 heads

**Solution**

If three coins are tossed at a time, then possible outcomes are

HHH, HHT, HTH, THH, TTH, THT, HTT, TTT

(i) Number of outcomes for getting 2 heads = m = 3

If ${E}_{1}$ is the event of getting two heads,

Then, P(${E}_{1}$) =$\frac{m}{n}=\frac{3}{8}$

Hence, probability of getting two heads is $\frac{3}{8}$ .

(ii) Number of outcomes for getting atleast 2 heads = m = 4

If ${E}_{2}$ is the event of getting atleast two heads,

Then, P(${E}_{2}$) = $\frac{m}{n}=\frac{4}{8}=\frac{1}{2}$

Hence, probability of getting atleast two heads is $\frac{1}{2}$ .

(iii) Number of outcomes of getting almost 2 heads = m = 7

If ${E}_{3}$ is the event of getting almost two heads,

Then, P(${E}_{3}$) =$\frac{m}{n}=\frac{7}{8}$

Hence, probability of getting almost two heads is $\frac{7}{8}$ .

**Example 5**One card is drawn from a well shuffled deck of 52 cards. Find the probability of drawing:

(i) a face card

(ii) ‘10’ of a red suit

(iii) an ace

(iv) ‘5’ of diamond

(v) a black queen

**Solution**

If a card is drawn randomly from a well shuffled pack of 52 cards, then total number of possible outcomes = n = 52

(i) Number of face cards in a pack of playing cards

=

*m*= 12

If ${E}_{1}$ is the event of drawing a face cad,

Then, P(${E}_{1}$) =$\frac{m}{n}=\frac{12}{52}=\frac{3}{13}$

Hence, probability of drawing a face card is 3/13.

(ii) There are two ‘10’ of red suit one in diamond and one is heart.

$\therefore $ Number of favourable outcomes = *m* = 2

If ${E}_{2}$ is the event of drawing ‘10 of a red suit,

Then P(${E}_{2}$) =$\frac{m}{n}=\frac{2}{52}=\frac{1}{26}$

Hence, probability of drawing ‘10’ of red suit is $\frac{1}{26}$ .

(iii) There are 4 aces in a deck.

Number of favourable outcomes = *m* = 4

If ${E}_{3}$ is the event of drawing an ace,

Then P(${E}_{3}$) =$\frac{m}{n}=\frac{4}{52}=\frac{1}{13}$

Hence, probability of drawing an ace is $\frac{1}{13}$ .

(iv) There is only one ‘5’ in diamond.

Number of favourable outcomes = *m* = 1

If ${E}_{4}$ is the event of drawing ‘5’ of diamond,

Then, P(${E}_{4}$) = $\frac{m}{n}=\frac{1}{52}$

Hence, probability of drawing ‘5’ of diamond is $\frac{1}{52}$.

(v) There are two black queens in a deck.

Number of favourable outcomes = *m* = 2

If ${E}_{5}$ is the event of drawing a black queen,

Then, P(${E}_{5}$) =$\frac{m}{n}=\frac{2}{52}=\frac{1}{26}$

Hence, probability of drawing a black queen =$\frac{1}{26}$ .