# Basic Theory of Probability, Introduction to Probability

## Probability — A Theoretical Approach

Let us consider the following situation:
Suppose a coin is tossed at random.

When we speak of a coin, we assume it to be ‘fair’, that is, it is symmetrical so that there is no reason for it to come down more often on one side than the other.
We call this property of the coin as being ‘unbiased’. By the phrase ‘random toss’, we mean that the coin is allowed to fall freely without any bias or interference.

We know, in advance, that the coin can only land in one of two possible ways — either head up or tail up (we dismiss the possibility of its ‘landing’ on its edge, which may be possible, for example, if it falls on sand). We can reasonably assume that each outcome, head or tail, is as likely to occur as the other. We refer to this by saying that the outcomes head and tail, are equally likely.
For another example of equally likely outcomes, suppose we throw a die once. For us, a die will always mean a fair die. What are the possible outcomes?
They are 1, 2, 3, 4, 5, 6. Each number has the same possibility of showing up. So the equally likely outcomes of throwing a die are 1, 2, 3, 4, 5 and 6.
Are the outcomes of every experiment equally likely? Let us see.
Suppose that a bag contains 4 red balls and 1 blue ball, and you draw a ball without looking into the bag. What are the outcomes? Are the outcomes — a red ball and a blue ball equally likely? Since there are 4 red balls and only one blue ball, you would agree that you are more likely to get a red ball than a blue ball. So, the outcomes (a red ball or a blue ball) are not equally likely. However, the outcome of drawing a ball of any colour from the bag is equally likely. So, all experiments do not necessarily have equally likely outcomes.
However, in this chapter, from now on, we will assume that all the experiments have equally likely outcomes.
In Class IX, we defined the experimental or empirical probability P(E) of an event E as

The empirical interpretation of probability can be applied to every event associated with an experiment which can be repeated a large number of times. The requirement of repeating an experiment has some limitations, as it may be very expensive or unfeasible in many situations. Of course, it worked well in coin tossing or die throwing experiments. But how about repeating the experiment of launching a satellite in order to compute the empirical probability of its failure during launching, or the repetition of the phenomenon of an earthquake to compute the empirical probability of a multi-storeyed building getting destroyed in an earthquake?
In experiments where we are prepared to make certain assumptions, the repetition of an experiment can be avoided, as the assumptions help in directly calculating the exact (theoretical) probability. The assumption of equally likely outcomes (which is valid in many experiments, as in the two examples above, of a coin and of a die) is one such assumption that leads us to the following definition of probability of an event.
The theoretical probability (also called classical probability) of an event E, written as P(E), is defined as

where we assume that the outcomes of the experiment are equally likely.
We will briefly refer to theoretical probability as probability.
This definition of probability was given by Pierre Simon Laplace in 1795.

Probability theory had its origin in the 16th century when an Italian physician and mathematician J.Cardan wrote the first book on the subject, The Book on Games of Chance.
Since its inception, the study of probability has attracted the attention of great mathematicians. James Bernoulli (1654 — 1705), A. de Moivre (1667 — 1754), and Pierre Simon Laplace are among those who made significant contributions to this field. Laplace’s Theorie Analytique des Probabilités, 1812, is considered to be the greatest contribution by a single person to the theory of probability. In recent years, probability has been used extensively in many areas such as biology, economics, genetics, physics, sociology etc.

Pierre Simon Laplace (1749-1827)

## Types of Experiments

1. Deterministic Experiment
Those experiments, which when repeated under identical conditions produce the same result or outcome are known as deterministic experiment,
2. Probabilistic/Random Experiment
Those experiments, which when repeated under identical conditions, do not produce the same outcome every time but the outcome in a trial is one of the several possible outcomes, called random experiment.

## Important Definitions

(i). Trial Let a random experiment, be repeated under identical conditions, then the experiment is called a Trial.
(ii). Sample Space The set of all possible outcomes of an experiment is called the sample space of the experiment and it is denoted by S.
(iii). Event A subset of the sample space associated with a random experiment is called event or case.
(iv). Sample Points The outcomes of an experiment is called the sample point.
(v). Certain Event An event which must occur, whatever be the outcomes, is called a certain or sure event.
(vi). Impossible Event An event which cannot occur in a particular random experiment, is called an impossible event.
(vii). Elementary Event An event certaining only one sample point is called elementary event or indeconnposable events.
(viii). Favourable Event Let S be the samplespace associated with a random experiment and let E⊂S. Then, the elementary events belonging to E are known as the favourable event to E.
(ix). Compound Events An event certaining more than one sample point is called compound events or decomposable events.

## Introduction

In earlier classes, we studied about the concept of probability as a measure of uncertainty of various phenomenon. We have obtained the probability of getting an even number in throwing a die as 3/6 i.e., ½. Here the total possible outcomes are 1,2,3,4,5 and 6 (six in number).
The outcomes in favour of the event of ‘getting an even number’ are 2,4,6 (i.e., three in number). In general, to obtain the probability of an event, we find the ratio of the
number of outcomes favourable to the event, to the total
number of equally likely outcomes. This theory of probability is known as classical theory of probability.

In Class IX, we learnt to find the probability on the basis of observations and collected data. This is called statistical approach of probability.
Both the theories have some serious difficulties. For instance, these theories can not be applied to the activities/experiments which have infinite number of outcomes. In classical theory we assume all the outcomes to be equally likely. Recall that the outcomes are called equally likely when we have no reason to believe that one is more likely to occur than the other. In other words, we assume that all outcomes have equal chance (probability) to occur. Thus, to define probability, we used equally likely or equally probable outcomes. This is logically not a correct definition. Thus, another theory of probability was developed by A.N. Kolmogorov, a Russian mathematician, in 1933. He laid down some axioms to interpret probability, in his book ‘Foundation of Probability’ published in 1933. In this Chapter, we will study about this approach called axiomatic approach of probability. To understand this approach we must know about few basic terms viz, random experiment, sample space, events, etc. Let us learn about these all, in what follows next.

## Random Experiments

In our day to day life, we perform many activities which have a fixed result no matter any number of times they are repeated. For example given any triangle, without knowing
the three angles, we can definitely say that the sum of measure of angles is 180°.
We also perform many experimental activities, where the result may not be same, when they are repeated under identical conditions. For example, when a coin is tossed it may turn up a head or a tail, but we are not sure which one of these results will actually be obtained. Such experiments are called random experiments.
An experiment is called random experiment if it satisfies the following two conditions:
(i). It has more than one possible outcome.
(ii). It is not possible to predict the outcome in advance.
Check whether the experiment of tossing a die is random or not?
In this chapter, we shall refer the random experiment by experiment only unless stated otherwise.

## Probability Simple-Formula

The concepts of probability play an important role in many problems of everyday life. It has very extensive applications in the development of physical sciences, economics, commerce, etc.
Probability of an Event:
If the sample space S is finite, then the probability P(E) of the event is given by:

If there are n elementary events associated with a random experiment and m of them are favourable to an event A, then the probability of happening or occurrence of A, denoted by P(A),
is given by
P(A)=mn= Number of favourable cases/Total number of possible cases.

## Impossible and Sure Event

Null set ‘Ø’ and Sample space ‘S’, are also subsets
of S. For example, on throwing of a dice event.
S={1,2,3,4,5,6} is sure to happen, hence its probability will be 1, And probability of getting any number other than {1,2,3,4,5,6} on its face is impossible to occur, hence its probability will be zero.
Also, 0≤P(E)≤1

## Outcomes and sample space

A possible result of a random experiment is called its outcome.
Consider the experiment of rolling a die. The outcomes of this experiment are 1, 2, 3, 4, 5, or 6, if we are interested in the number of dots on the upper face of the die.
The set of outcomes {1, 2, 3, 4, 5, 6} is called the sample space of the experiment.
Thus, the set of all possible outcomes of a random experiment is called the sample space associated with the experiment. Sample space is denoted by the symbol S.
Each element of the sample space is called a sample point. In other words, each outcome of the random experiment is also called sample point.
For example:
On throwing a die, the sample space
S={1,2,3,4,5,6} and n(S)=6
E={2,3,5} is a subset of S. So E is an event and n(E)=3.
The event E={2,3,5} is also expressed as the event of getting a prime number in throwing a die.

## Event

We have studied about random experiment and sample space associated with an experiment. The sample space serves as an universal set for all questions concerned with the experiment.
Consider the experiment of tossing a coin two times. An associated sample space is S={HH,HT,TH,TT}.
Now suppose that we are interested in those outcomes which correspond to the occurrence of exactly one head. We find that HT and TH are the only elements of S corresponding to the occurrence of this happening (event). These two elements form the set E={HT,TH}
We know that the set E is a subset of the sample space S. Similarly, we find the following correspondence between events and subsets of S.
Description of events Corresponding subset of ‘S’
Number of tails is exactly 2, A={TT}
Number of tails is atleast one, B={HT,TH,TT}
Number of heads is atmost one, C={HT,TH,TT}
Second toss is not head, D={HT,TT}
Number of tails is atmost two, S={HH,HT,TH,TT}
Number of tails is more than two, Ø
The above discussion suggests that a subset of sample space is associated with an event and an event is associated with a subset of sample space. In the light of this we define an event as follows.
Definition: Any subset E of a sample space S is called an event.

## Occurrence of an event

Consider the experiment of throwing a die. Let E denotes the event “ number less than 4 appears”. If actually ‘I’ had appeared on the die then we say that event E has occurred. As a matter of fact if outcomes are 2 or 3, we say that event E has occurred.
Thus, the event E of a sample space S is said to have occurred if the outcome ω of the experiment is such that ω∈E. If the outcome ω is such that ω∉E, we say that the event E has not occurred.

## Examples on finding the probability of respective events

In what follows, S is the sample space of the experiment in question and E is the eventof interest. n(S) is the number of elements in the sample space S and n(E) is the number of elements in the event E.

Ex1. Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to answer a true -false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.
(i) It is not an equally likely event, as it depends on various factors such as whether the car will start or not. And factors for both the conditions are not the same.
(ii) It is not an equally likely event, as it depends on the player’s ability and there is no information given about that.
(iii) It is an equally likely event.
(iv) It is an equally likely event.

Ex2. Fill in the blanks:
(i) Probability of a sure event is ….
(ii) Probability of an impossible event is ….
(iii) The probability of an event (other than sure and impossible event) lies between ….
(iv) Every elementary event associated to a random experiment has … probability.
(v) Probability of an event A + Probability of event ‘not A’ =⋯.
(vi) Sum of the probabilities of each outcome m an experiment is ….
Sol:
(i) 1, ∵P(sure event)=1
(ii) 0, ∵P(impossible event)=0
(iii) 0P(E)+P(Ē)=1
(vi) 1

Ex3. A bag contains lemon flavored candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
(i) an orange flavored candy?
(ii) a lemon flavored candy?
Sol:
(i) The bag contains lemon flavored candies only. So, the event that malini will take out an orange flavored candy is an impossible event. Since, probability of impossible event is 0, P(an orange flavored candy) =0
(ii) The bag contains lemon flavored candies only. So, the event that malini will take out a lemon flavored candy is sure event. Since probability of sure event is 1, P(a lemon flavored candy) =1

Ex4. Which of the following cannot be the probability of an event?
(A) ⅔ (B) -1.5 (C) 15% (D) 0.7
Probability of an event (E) is always greater than or equal to 0. Also, it is always less than or equal to one. This implies that the probability of an event cannot be negative or greater than 1. Therefore, out of these alternatives, -1.5 cannot be a probability of an event.
Hence, (B)

Ex5. Which of these numbers cannot be a probability?
a) -0.001 b) 0.5 c) 1.001 d) 0 e) 1 f) 20%
Solution:
A probability is always greater than or equal to 0 and less than or equal to 1, hence only a) and c) above cannot represent probabilities: -0.001 is less than 0 and 1.001 is greater than 1.

Ex6. Which of the following cannot be the probability of an event? (i) ⅗ (ii) 2.7 (iii) 43% (iv) -0.6 (v) -3.2 (vi) 0.35
Solution:
The probability of an event lies between ‘0 and 1’ i.e.

0≤P(E)≤1
0≤P(E)≤100%

Hence, (i), (iii), (vi) can be three values of probability.
Hence, (ii), (iv), (v) can’t be three values of probability.

Ex7. Which of the following cannot be the probability of an event?
(i) 3/7 (ii) 0.82 (iii) 37% (iv) -2.4
Sol:
probability of an event E is 0≤P(E)≤1
0≤{(i),(ii),(iii)}≤1
Therefore. Either (i),(ii) or (iii) can be a probability of an event.
(iv) Since -2.4< 0
Therefore, (iv) cannot be a probability of an event.

Ex8. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?
When we toss a coin, the possible outcomes are only two, head or tail, which are equally likely outcomes. Therefore, the result of an individual toss is completely unpredictable.

Ex9. A child has a die whose six faces shows the letters as given below:

[A][B][C][D][E][A]

The die is thrown once. What is the probability of getting (i) A? (ii) D?
Total number of possible outcomes on the dice =6
(i) Total number of faces having A on it =2
P(getting A)=2/6=⅓

(ii) Total number of faces having D on it =1
P(getting D)=⅙

Ex10. Which of the following arguments are correct and which are not correct? Give reasons for your answer.
(i) If two coins are tossed simultaneously there are three possible outcomes–two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is ⅓.
(ii) If a die is thrown, there are two possible outcomes-an odd number or an even number. Therefore, the probability of getting an odd number is ½.
(i) Incorrect
When two coins are tossed, the possible outcomes are (H, H), (H, T), (T, H), and (T, T). It can be observed that there can be one of each in two possible ways — (H, T), (T, H).
Therefore, the probability of getting two heads is ¼, the probability of getting two tails is ¼, and the probability of getting one of each is ½. It can be observed that for each outcome, the probability is not ⅓.
(ii) Correct
When a dice is thrown, the possible outcomes are 1, 2, 3, 4, 5, and 6. Out of these, 1, 3, 5 are odd and 2,4, 6 are even numbers. Therefore, the probability of getting an odd number is ½.

Ex11. Five cards — the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put a side, what is the probability that the second card picked up is
a. an ace? b. a queen?
Sol:
Total number of possible outcomes n(S)=5 (5 cards)
(i) E→ event of getting a queen, n(E)=1

(ii) If queen is drawn & put aside, total number of remaining cards n(S)=4
(a) E→ event of getting an ace, n(E)=1
P(E)=¼

(b) E→ event of getting another queen, n(E)=0
P(E)=0

We know it is an impossible event.

Ex12. In a lottery there are 10 prizes and 25 blanks. What is the probability of getting a prize?
Sol:
Total number of possible outcomes n(S)=35 {10 prizes, 25 blanks)
E→ event of getting prize
Number of favorable outcomes n(E)=10 {10 prizes)

Ex13. In a class, there are 18 girls and 16 boys. The class teacher wants to choose one pupil for class monitor. What she does, she writes the name of each pupil on a card and puts them into a basket and mixes thoroughly. A child is asked to pick one card from the basket. What is the probability that the name written on the card is:
(i) the name of a girl (ii) the name of a boy
Sol:
Total number of possible outcomes n(S)=18+16=34
(i) E→ event of getting girl name
Number of favorable outcomes n(E)=18 (18 girls)

(ii) E→ event of getting boy name
Number of favorable outcomes n(E)=16 (16 boys)

Ex14. Twelve defective pens are accidently mixed with 132 good ones. It is not possible to just look at pen and tell whether or not it is defective. one pen is taken out at random from this lot. Determine the probability that the pen taken out is good one.
Sol:
Number of good pens =132
Number of defective pens =12
Total number of possible outcomes n(S)=132+12=144 (total number of pens}
E→ event of getting a good pen.
Number of favorable outcomes n(E)=132 (132 good pens}

Ex15. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see the given figure),

and these are equally likely outcomes. What is the probability that it will point at
(i) 8? (ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?
Total number of possible outcomes =8
(i) Probability of getting 8=⅛
(ii) Total number of odd numbers on spinner =4
probability or getting an odd number=4/8=½
(iii) The numbers greater than 2 are 3, 4, 5, 6, 7, and 8. Therefore, total numbers greater than 2=6
Probability of getting a number greater than 2 =6/8=¾
(iv) The numbers less than 9 are 1, 2, 3, 4, 6, 7, and 8.
Therefore, total numbers less than 9 is equal to 8.
Probability of getting a number less than 9 =8/8=1

Multiple Choice: Identify the letter of the choice that best completes the statement or answers the question.
Use The Following To Answer Questions 1-2:
Greg spins the spinner twice.

1. What is the probability that it will land on an even number both times?
a. ⅛ b. 1/16 c. ¼ d. ½
correct: c, solution:
There are 16 possible outcomes, 4 of which have two even numbers.
(2, 2), (2, 3), (2, 4), (2, 5)
(3, 2), (3, 3), (3, 4), (3, 5)
(4, 2), (4, 3), (4, 4), (4, 5)
(5, 2), (5, 3), (5, 4), (5, 5)

The probability of spinning two even numbers is 4/16=¼. A shortway gives
2/4×2/4=½×½=¼

2. What is the probability that the spinner will land on 5 on the first spin and 2 on the second spin?
a. 1/16 b. ⅛ c. 3/16 d. ¼
correct: a, solution:
There are 16 possible outcomes. Of these, only 1 is spinning 5 first and 2 second.

(2, 2), (2, 3), (2, 4), (2, 5)
(3, 2), (3, 3), (3, 4), (3, 5)
(4, 2), (4, 3), (4, 4), (4, 5)
(5, 2), (5, 3), (5, 4), (5, 5)

The probability of spinning 5, then 2, is 1/16.

Use The Following To Answer Questions 3-4:
Mrs. Liang spins each spinner one time.

3. What is the probability that the first spinner will land on an odd number and the second spinner will land on a vowel?
a. 1/12 b. ¼ c. ⅓ d. ⅙
correct: d, solution:
There are 12 possible outcomes. Of these, 2 have an odd number first and an A second.
(6, A), (6, B), (6, C)
(7, A), (7, B), (7, C)
(8, A), (8, B), (8, C)
(9, A), (9, B), (9, C)

The probability of spinning an odd number, then a vowel, is 2/12=⅙.

4. What is the probability that the first spinner will land on 7 and the second spinner will land on C?
a. ⅙ b. ⅓ c. ¼ d. 1/12
correct: d, solution:
There are 12 possible outcomes. Of these, only 1 is a 7 first and a C second.

(6, A), (6, B), (6, C)
(7, A), (7, B), (7, C)
(8, A), (8, B), (8, C)
(9, A), (9, B), (9, C)

The probability of spinning a 7, then a C, is 1/12.

Use The Following To Answer Questions 5-6:
Jared is going to perform an experiment in which he spins each spinner once.

5. What is the probability that the first spinner will land on A, the second spinner will land on an even number, and the third spinner will land on Blue?
a. ⅛ b. 1/12 c. 1/24 d. ¼
correct: b, solution:
There are 24 possible outcomes. Of these, 2 have an A first, an even number second, and Blue third.
So, the probability of spinning “A, an even number, and Blue” is 2/24=1/12. A shortway gives

6. What is the probability that the first spinner will land on B, the second spinner will land on 3, and the third spinner will land on Green?
a. ¼ b. ⅛ c. 1/12 d. 1/24
correct: d, solution:
There are 24 possible outcomes. Of these, 1 has B first, 3 second, and Green third.
So, the probability of spinning “A, 3, and Green” is 1/24.