Chapter Overview

Introduction

• The terms 'probably', 'in all likelihood', 'chance', 'odds in favor', and 'odds against' are now quite common and originate from a mathematical field known as Probability. Recently, probability has evolved into a comprehensive subject and has become a crucial component of statistics. The theories of Hypothesis Testing and Estimation are founded on probability.
• Interestingly, the initial application of probability was undertaken by a group of European mathematicians around three hundred years ago to improve their chances in various gambling games. Subsequently, the theory of probability was further developed by notable figures such as Abraham De Moivre and Pierre-Simon Laplace from France, Reverend Thomas Bayes and R. A. Fisher from England, as well as Chebyshev, Markov, Khinchin, Kolmogorov from Russia, alongside many other distinguished mathematicians and statisticians.
• Probability can be broadly categorized into two types: Subjective Probability and Objective Probability. Subjective Probability relies on personal judgment and experience, making it susceptible to the individual's beliefs, attitudes, and biases. Nonetheless, in uncertain situations, it proves useful and is applied in decision-making management. However, this discussion will focus solely on Objective Probability.

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Random Experiment

To build a solid understanding of probability, it's important to become familiar with some key terms. 

Key Terms in Probability

• Experiment: An experiment refers to a process or activity that produces specific outcomes. 
• Random Experiment:. random experiment is one where the results are determined by chance. For instance, when you toss a coin, you get two possible outcomes: Head (H) or Tail (T). It's impossible to predict in advance which one will come up, making it a random experiment. Other examples include rolling dice, drawing items from a mixed box, or picking cards from a shuffled deck. 
• Events: Events are the outcomes or results of a random experiment. Sometimes, events can be a combination of multiple outcomes. There are two types of events: 
• (i) Simple or Elementary Events: These are events that cannot be further divided. For example, when you toss a coin, you get two simple events: Head and Tail. 
• (ii) Composite or Compound Events: These are events that can be broken down into smaller events. For instance, getting a head when a coin is tossed twice is a composite event because it can be split into the events HT and TH, which are both simple events. 
• Mutually Exclusive Events:. set of events is considered mutually exclusive if only one of them can occur at the same time. For example, when you toss a coin, you get two mutually exclusive events: Head and Tail. If one occurs, the other cannot. 
• Exhaustive Events: Events are said to be exhaustive if at least one of them must occur. For instance, when tossing a coin, the events Head and Tail are exhaustive because these are the only possible outcomes. 
• Equally Likely Events: Events are considered equally likely when there is no reason to believe that one event will occur more frequently than another. For example, when tossing a coin, Head and Tail are equally likely events because there is no basis to assume that one will occur more often than the other. 

 Classical Definition of Probability or A Priori Definition

Let us consider a random experiment that result in n finite elementary events, which are assumed to be equally likely. We next assume that out of these n events, n_A ( ≤ n) events are favourable to an event A. Then the probability of occurrence of the event A is defined as the ratio of the number of events favourable to A to the total number of events. Denoting this by P(A), we have

However if instead of considering all elementary events, we focus our attention to only those composite events, which are mutually exclusive, exhaustive and equally likely and if m(≤ n) denotes such events and is furthermore m_A(≤n_A) denotes the no. of mutually exclusive, exhaustive and equally likely events favourable to A, then we have

For this definition of probability, we are indebted to Bernoulli and Laplace. This definition is also termed as a priori definition because probability of the event A is defined on the basis of prior knowledge.
This classical definition of probability has the following demerits or limitations:
(i) It is applicable only when the total no. of events is finite.
(ii) It can be used only when the events are equally likely or equi-probable. This assumption is made well before the experiment is performed.
(iii) This definition has only a limited field of application like coin tossing, dice throwing, drawing cards etc. where the possible events are known well in advance. In the field of uncertainty or where no prior knowledge is provided, this definition is inapplicable.
In connection with classical definition of probability, we may note the following points:

ILLUSTRATIONS:

Example 15.1: A coin is tossed three times. What is the probability of getting:
(i) 2 heads
(ii) at least 2 heads.

Solution: When a coin is tossed three times, first we need enumerate all the elementary events. This can be done using 'Tree diagram' as shown below:

Hence the elementary events are
HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
Thus the number of elementary events (n) is 8.

(i) Out of these 8 outcomes, 2 heads occur in three cases namely HHT, HTH and THH. If we denote the occurrence of 2 heads by the event A and if assume that the coin as well as performer of the experiment is unbiased then this assumption ensures that all the eight elementary events are equally likely. Then by the classical definition of probability, we have

(ii) Let B denote occurrence of at least 2 heads i.e. 2 heads or 3 heads. Since 2 heads occur in 3 cases and 3 heads occur in only 1 case, B occurs in 3 + 1 or 4 cases. By the classical definition of probability,

P(B) = 4  / 8 
= 0.50

Example 15.2: A dice is rolled twice. What is the probability of getting a difference of 2 points?

Solution: If an experiment results in p outcomes and if the experiment is repeated q times, then the total number of outcomes is pq. In the present case, since a dice results in 6 outcomes and the dice is rolled twice, total no. of outcomes or elementary events is 62 or 36. We assume that the dice is unbiased which ensures that all these 36 elementary events are equally likely. Now a difference of 2 points in the uppermost faces of the dice thrown twice can occur in the following cases:

Thus denoting the event of getting a difference of 2 points by A, we find that the no. of outcomes favourable to A, from the above table, is 8. By classical definition of probability, we get

P(A)= 8 /36

= 2/9

Example 15.3: Two dice are thrown simultaneously. Find the probability that the sum of points on the two dice would be 7 or more.

Solution: If two dice are thrown then, as explained in the last problem, total no. of elementary events is 62 or 36. Now a total of 7 or more i.e. 7 or 8 or 9 or 10 or 11 or 12 can occur only in the following combinations:

SUM = 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

SUM = 8: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2)

SUM = 9: (3, 6), (4, 5), (5, 4), (6, 3)

SUM = 10: (4, 6), (5, 5), (6, 4)

SUM = 11: (5, 6), (6, 5)

SUM = 12: (6, 6)

Thus the no. of favourable outcomes is 21. Letting A stand for getting a total of 7 points or more, we have

P(A) = 21 / 36 
= 7 / 12

Example 15.4: What is the chance of picking a spade or an ace not of spade from a pack of 52 cards?

Solution: A pack of 52 cards contain 13 Spades, 13 Hearts, 13 Clubs and 13 Diamonds. Each of these groups of 13 cards has an ace. Hence the total number of elementary events is 52 out of which 13 + 3 or 16 are favourable to the event A representing picking a Spade or an ace not of Spade. Thus we have

P(A) = 16 / 52 
= 4/13

P(A∪B) = P(A) + P(B) – P(A∩B)

= 13 /52 = 3/52 – 0 = 16 / 52 = 4/13

Example 15.5: Find the probability that a four digit number comprising the digits 2, 5, 6 and 7 would be divisible by 4.

Solution: Since there are four digits, all distinct, the total number of four digit numbers that can be formed without any restriction is 4! or 4 × 3 × 2 × 1 or 24. Now a four digit number would be divisible by 4 if the number formed by the last two digits is divisible by 4. This could happen when the four digit number ends with 52 or 56 or 72 or 76. If we fix the last two digits by 52, and then the 1st two places of the four digit number can be filled up using the remaining 2 digits in 2! or 2 ways. Thus there are 2 four digit numbers that end with 52. Proceeding in this manner, we find that the number of four digit numbers that are divisible by 4 is 4 × 2 or 8. If (A) denotes the event that any four digit number using the given digits would be divisible by 4, then we have

P(A) = 8 / 24 = 1 / 3

Example 15.6: A committee of 7 members is to be formed from a group comprising 8 gentlemen and 5 ladies. What is the probability that the committee would comprise:
(a) 2 ladies,
(b) at least 2 ladies.

Solution: Since there are altogether 8 + 5 or 13 persons, a committee comprising 7 members can be formed in

or 11 × 12 × 13 ways.

(a) When the committee is formed taking 2 ladies out of 5 ladies, the remaining (7–2) or 5 committee members are to be selected from 8 gentlemen. Now 2 out of 5 ladies can be selected in ⁵C₂ ways and 5 out of 8 gentlemen can be selected in ⁸C₅ ways. Thus if A denotes the event of having the committee with 2 ladies, then A can occur in ⁵C₂× ⁸C₅ or

(b) Since the minimum number of ladies is 2, we can have the following combinations:

Thus if B denotes the event of having at least two ladies in the committee, then B can occur in

i.e. 1568 ways.

Hence 

Relative Frequency Definition of Probability

Owing to the limitations of the classical definition of probability, there are cases when we consider the statistical definition of probability based on the concept of relative frequency. This definition of probability was first developed by the British mathematicians in connection with the survival probability of a group of people.

Let us consider a random experiment repeated a very good number of times, say n, under an identical set of conditions. We next assume that an event A occurs fA times. Then the limiting value of the ratio of fA to n as n tends to infinity is defined as the probability of A.

This statistical definition is applicable if the above limit exists and tends to a finite value.

Example 15.7: The following data relate to the distribution of wages of a group of workers:

If a worker is selected at random from the entire group of workers, what is the probability that 
(a) his wage would be less than ₹ 50? 
(b) his wage would be less than ₹ 80? 
(c) his wage would be more than ₹ 100? 
(d) his wages would be between ₹ 70 and ₹ 100?

Solution: As there are altogether 150 workers, n = 150. 
(a) Since there is no worker with wage less than ₹ 50, the probability that the wage of a randomly selected worker would be less than ₹50 is P(A) = 0 / 150 = 0

(b) Since there are (15+23+36) or 74 worker having wages less than ` 80 out of a group of 150 workers, the probability that the wage of a worker, selected at random from the group, would be less than ₹ 80 is

(c) There are (12+5) or 17 workers with wages more than ₹100. Thus the probability of finding a worker, selected at random, with wage more than ₹ 100 is

(d) There are (36 + 42 + 17) or 95 workers with wages in between ₹ 70 and ₹ 100. Thus

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Operations on Events - Set Theoretic Approach to Probability

Applying the concept of set theory, we can give a new dimension to the classical definition of probability. A sample space may be defined as a non-empty set containing all the elementary events of a random experiment as sample points. A sample space is denoted by S or  . An event A may be defined as a non-empty subset of S. This is shown in Figure 15.1

As for example, if a dice is rolled once than the sample space is given by
S = {1, 2, 3, 4, 5, 6}.
Next, if we define the events A, B and C such that
A = {x: x is an even no. of points in S}
B = {x: x is an odd no. of points in S}
C = {x: x is a multiple of 3 points in S}
Then, it is quite obvious that
A = {2, 4, 6}, B = {1, 3, 5} and C = {3, 6}.
The classical definition of probability may be defined in the following way.
Let us consider a finite sample space S i.e. a sample space with a finite no. of sample points, n (S). We assume that all these sample points are equally likely. If an event A which is a subset of S, contains n (A) sample points, then the probability of A is defined as the ratio of the number of sample points in A to the total number of sample points in S. i.e.

Union and Intersection of Two Events

Union of two events A and B is defined as a set of events containing all the sample points of event A or event B or both the events. This is shown in Figure 15.2 and

where x denotes the sample points.

In the above example, we have A∪C = {2, 3, 4, 6}
and A ∪ B = {1, 2, 3, 4, 5, 6}.
The intersection of two events A and B may be defined as the set containing all the sample points that are common to both the events A and B. This is shown in Figure 15.2. we have

In the above example, A ∩ B = Φ

Since the intersection of the events A and B is a null set Φ , it is obvious that A and B are mutually exclusive events as they cannot occur simultaneously.
The difference of two events A and B, to be denoted by A – B, may be defined as the set of sample points present in set A but not in B. i.e.

Similarly,

In the above examples,

And A – C = {2, 4}.

This is shown in Figure 15.3.

The complement of an event A may be defined as the difference between the sample space S and the event A. i.e.

In the above example A’= S – A
= {1, 3, 5}
Figure 15.4 depicts A’

Showing A’
Now we are in a position to redefine some of the terms we have already discussed in this section.
Two events A and B are mutually exclusive if P (A ∩B) = 0 or more precisely, ....(15.9)

……….(15.10)

Similarly three events A, B and C are mutually exclusive if

Example 15.8: Three events A, B and C are mutually exclusive, exhaustive and equally likely. What is the probability of the complementary event of A?

Solution: Since A, B and C are mutually exclusive, we have

Axiomatic or Modern Definition of Probability

Let us consider a sample space S in connection with a random experiment and let A be an event defined on the sample space S i.e. A  S. Then a real valued function P defined on S is known as a probability measure and P(A) is defined as the probability of A if P satisfies the following axioms:

Addition Theorems or Theorems on Total Probability

Theorem 1 For any two mutually exclusive events A and B, the probability that either A or B occurs is given by the sum of individual probabilities of A and B.

This is illustrated in the following example.

Example 15.9: A number is selected from the first 25 natural numbers. What is the probability that it would be divisible by 4 or 7?

Solution: Let A be the event that the number selected would be divisible by 4 and B, the event that the selected number would be divisible by 7. Then AUB denotes the event that the number would be divisible by 4 or 7. Next we note that A = {4, 8, 12, 16, 20, 24} and B = {7, 14, 21} whereas S = {1, 2, 3, ……... 25}. Since A ∩ B = Φ , the two events A and B are mutually exclusive and as such we have

Thus from (1), we have

Hence the probability that the selected number would be divisible by 4 or 7 is 9/25 or 0.36 

Example 15.10: A coin is tossed thrice. What is the probability of getting 2 or more heads? 

Solution: If a coin is tossed three times, then we have the following sample space.

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} 2 or more heads imply 2 or 3 heads. If A and B denote the events of occurrence of 2 and 3 heads respectively, then we find that

A = {HHT, HTH, THH} and B = {HHH}

and

As A and B are mutually exclusive, the probability of getting 2 or more heads is

Theorem 2 For any K( ≥ 2) mutually exclusive events A₁, A₂, A₃ …, A_K the probability that at least one of them occurs is given by the sum of the individual probabilities of the K events.

Obviously, this is an extension of Theorem 1.

Theorem 3 For any two events A and B, the probability that either A or B occurs is given by the sum of individual probabilities of A and B less the probability of simultaneous occurrence of the events A and B.

This theorem is stronger than Theorem 1 as we can derive Theorem 1 from Theorem 3 and not Theorem 3 from Theorem 1. For want of sufficient evidence, it is wiser to apply Theorem 3 for evaluating total probability of two events.

Example 15.11: A number is selected at random from the first 1000 natural numbers. What is the probability that it would be a multiple of 5 or 9?

Solution: Let A, B, A  ∪ B and A ∩ B denote the events that the selected number would be a multiple of 5, 9, 5 or 9 and both 5 and 9 i.e. LCM of 5 and 9 i.e. 45 respectively.

multiple of 5 since 1000 = 5 × 200

multiple of 9 = 9 × 111 + 1

Both 5 and 9 = 45 × 22 + 10,

it is obvious that

Hence the probability that the selected number would be a multiple of 4 or 9 is given by

Example 15.12: The probability that an Accountant's job applicant has a B. Com. Degree is 0.85, that he is a CA is 0.30 and that he is both B. Com. and CA is 0.25 out of 500 applicants, how many would be B. Com. or CA?

Solution: Let the event that the applicant is a B. Com. be denoted by B and that he is a CA be denoted by C Then as given,

P(B) = 0.85, P(C) = 0.30 and P(B ∩ C) = 0.25

The probability that an applicant is B. Com. or CA is given by

P(B ∪ C) = P(B) + P(C) – P(B ∩ C)

= 0.85 + 0.30 – 0.25

Expected frequency = N × P (B ∪ C)

Expected frequency = 500 × 0.90 = 450

Example 15.13: If P(A–B) = 1/5, P(A) = 1/3 and P (B) = 1/2, what is the probability that out of the two events A and B, only B would occur?

Solution: A glance at Figure 15.3 suggests that

Also (15.21) and (15.22) describe the probabilities of occurrence of the event only A and only B respectively.

The probability that the event B only would occur

Theorem 4 For any three events A, B and C, the probability that at least one of the events occurs is given by

Following is an application of this theorem.

Example 15.14: There are three persons A, B and C having different ages. The probability that A survives another 5 years is 0.80, B survives another 5 years is 0.60 and C survives another 5 years is 0.50. The probabilities that A and B survive another 5 years is 0.46, B and C survive another 5 years is 0.32 and A and C survive another 5 years 0.48. The probability that all these three persons survive another 5 years is 0.26. Find the probability that at least one of them survives another 5 years.

Solution: As given P(A) = 0.80, P(B) = 0.60, P(C) = 0.50,

P(A ∩ B) = 0.46, P(B ∩ C) = 0.32, P(A ∩ C) = 0.48 and

P(A ∩ B ∩ C) = 0.26

The probability that at least one of them survives another 5 years in given by

P(A ∪ B ∪ C)

= P(A) + P(B) + P(C) – P(A ∩ B) – P(A ∩ C) – P(B ∩ C)+ P(A ∩ B ∩ C) ……. (15.23)

= 0.80 + 0.60 + 0.50 – 0.46 – 0.32 – 0.48 + 0.26

= 0.90

Expected Frequency = N X P(A ∪ B ∪ C) =500 X 0.90= 450

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Conditional Probability and Compound Theorem of Probability

Compound Probability or Joint Probability
The probability of an event, discussed so far, is technically known as unconditional or marginal probability. However, there are situations that demand the probability of occurrence of more than one event. The probability of occurrence of two events A and B simultaneously is known as the Compound Probability or Joint Probability of the events A and B and is denoted by P(A ∩ B). In a similar manner, the probability of simultaneous occurrence of K events A₁, A₂, …. A_k, is denoted by P(A1 ∩ A2 ∩ …. ∩ Ak).

In case of compound probability of 2 events A and B, we may face two different situations. In the first case, if the occurrence of one event, say B, is influenced by the occurrence of another event A, then the two events A and B are known as dependent events. We use the notation P(B/A), to be read as 'probability of the event B given that the event A has already occurred (or 'the conditional probability of B given A) to suggest that another event B will happen if and only if the first event A has already happened. This is given by

Provided P(A) > 0 i.e. A is not an impossible event.

Similarly,

if P(B) > 0.
As an example if a box contains 5 red and 8 white balls and two successive draws of 2 balls are made from it without replacement then the probability of the event 'the second draw would result in 2 white balls given that the first draw has resulted in 2 Red balls' is an example of conditional probability since the drawings are made without replacement, the composition of the balls in the box changes and the occurrence of 2 white balls in the second draw (B₂) is dependent on the outcome of the first draw (R₂). This event may b denoted by P(B₂/R₂).

In the second scenario, if the occurrence of the second event B is not influenced by the occurrence of the first event A, then B is known to be independent of A. It also follows that in this case, A is also independent of B and A and B are known as mutually independent or just independent. In this case, we have

In the above example, if the balls are drawn with replacement, then the two events B2 and R2 are independent and we have

P(B₂ / R₂) = P(B₂)

(15.28) is the necessary and sufficient condition for the independence of two events. In a similar manner, three events A, B and C are known as independent if the following conditions hold :

It may be further noted that if two events A and B are independent, then the following pairs of events are also independent:
(i) A and B’
(ii) A’ and B
(iii) A’ and B’ ……… (15.30)

Theorems of Compound Probability
Theorem 5 For any two events A and B, the probability that A and B occur simultaneously is given by the product of the unconditional probability of A and the conditional probability of B given that A has already occurred

Theorem 6 For any three events A, B and C, the probability that they occur jointly is given by

which we have already discussed.

Example 15.15: Rupesh is known to hit a target in 5 out of 9 shots whereas David is known to hit the same target in 6 out of 11 shots. What is the probability that the target would be hit once they both try?

Solution: Let A denote the event that Rupesh hits the target and B, the event that David hits the target. Then as given,

The probability that the target would be hit is given by

Example 15.16: A pair of dice is thrown together and the sum of points of the two dice is noted to be 10. What is the probability that one of the two dice has shown the point 4?

Solution: Let A denote the event of getting 4 points on one of the two dice and B denote the event of getting a total of 10 points on the two dice. Then we have

Alternately The sample space for getting a total of 10 points when two dice are thrown simultaneously is given by
S = {(4, 6), (5, 5), (6, 4)}
Out of these 3 cases, we get 4 in 2 cases. Thus by the definition of probability, we have
P(B/A) = 2/ 3

Example 15.17: In a group of 20 males and 15 females, 12 males and 8 females are service holders. What is the probability that a person selected at random from the group is a service holder given that the selected person is a male?

Solution: Let S and M stand for service holder and male respectively. We are to evaluate P (S / M).

We note that (S ∩ M) represents the event of both service holder and male.

Thus 

Example 15.18: In connection with a random experiment, it is found that
P(A) = 2 /3 , P(B) 3 /5 = and P(A ∪ B) = 5 6
Evaluate the following probabilities:
(i) P(A/B) (ii) P(B/A) (iii) P(A’/ B) (iv) P(A/ B’) (v) P(A’/ B’)

Solution: P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ 5 /6 = 2/ 3 + 3/ 5 – P(A ∩ B)
⇒ P(A∩ B) = 2 / 3 + 3 / 5 – 5 / 6
= 13 / 30

Hence (i)

= 5 / 12

Example 15.19: The odds in favour of an event is 2 : 3 and the odds against another event is 3 : 7. Find the probability that only one of the two events occurs.

Solution: We denote the two events by A and B respectively. Then by (15.5) and (15.6), we have

As A and B are independent, P(A ∩ B) = P(A) ∩ P(B)

Probability that either only A occurs or only B occurs

Example 15.20: There are three boxes with the following compositions :

One ball is drawn from each box. What is the probability that they would be of the same colour?

Solution: Either the balls would be Blue or Red or White. Denoting Blue, Red and White balls by B, R and W respectively and the box by lower suffix, the required probability is

Example 15.21: Mr. Roy is selected for three separate posts. For the first post, there are three candidates, for the second, there are five candidates and for the third, there are 10 candidates. What is the probability that Mr. Roy would be selected?

Solution: Denoting the three posts by A, B and C respectively, we have

Example 15.22: The independent probabilities that the three sections of a costing department will encounter a computer error are 0.2, 0.3 and 0.1 per week respectively. What is the probability that there would be
(i) at least one computer error per week?
(ii) one and only one computer error per week?

Solution: Denoting the three sections by A, B and C respectively, the probabilities of encountering a computer error by these three sections are given by P(A) = 0.20, P(B) = 0.30 and P(C) = 0.10

(i) Probability that there would be at least one computer error per week.
= 1 – Probability of having no computer error in any at the three sections. 

(ii) Probability of having one and only one computer error per week

Example 15.23: A lot of 10 electronic components is known to include 3 defective parts. If a sample of 4 components is selected at random from the lot, what is the probability that this sample does not contain more than one defectives?

Solution: Denoting defective component and non-defective components by D and D’ respectively, we have the following situation :

Thus the required probability is given by

Example 15.24: There are two urns containing 5 red and 6 white balls and 3 red and 7 white balls respectively. If two balls are drawn from the first urn without replacement and transferred to the second urn and then a draw of another two balls is made from it, what is the probability that both the balls drawn are red?

Solution: Since two balls are transferred from the first urn containing 5 red and 6 white balls to the second urn containing 3 red and 7 white balls, we are to consider the following cases :
Case A : Both the balls transferred are red. In this case, the second urn contains 5 red and 7 white balls.
Case B : The two balls transferred are of different colours. Then the second urn contains 4 red and 8 white balls.
Case C : Both the balls transferred are white. Now the second urn contains 3 red and 9 white balls.

The required probability is given by 

Example 15.25: If 8 balls are distributed at random among three boxes, what is the probability that the first box would contain 3 balls?

Solution: The first ball can be distributed to the 1st box or 2nd box or 3rd box i.e. it can be distributed in 3 ways. Similarly, the second ball also can be distributed in 3 ways. Thus the first two balls can be distributed in 3² ways. Proceeding in this way, we find that 8 balls can be distributed to 3 boxes in 3⁸ ways which is the total number of elementary events.

Let A be the event that the first box contains 3 balls which implies that the remaining 5 balls must go to the remaining 2 boxes which, as we have already discussed, can be done in 2⁵ ways. Since 3 balls out of 8 balls can be selected in ⁸C₃ ways, the event can occur in ⁸C₃ x 2⁵ ways, thus we have

Example 15.26: There are 3 boxes with the following composition :
Box I : 7 Red + 5 White + 4 Blue balls
Box II : 5 Red + 6 White + 3 Blue balls
Box III : 4 Red + 3 White + 2 Blue balls
One of the boxes is selected at random and a ball is drawn from it. What is the probability that the drawn ball is red?

Solution: Let A denote the event that the drawn ball is red. Since any of the 3 boxes may be

drawn, we have P(B_I) = P(B_II) = P(B_III)= 1/ 3

Also P(R₁/B_II) = probability of drawing a red ball from the first box

= 7 / 16

Random Variable - Probability Distribution

A random variable or stochastic variable is a function defined on a sample space associated with a random experiment assuming any value from R and assigning a real number to each and every sample point of the random experiment. A random variable is denoted by a capital letter. For example, if a coin is tossed three times and if X denotes the number of heads, then X is a random variable. In this case, the sample space is given by
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
and we find that X = 0 if the sample point is TTT
X = 1 if the sample point is HTT, THT or TTH
X = 2 if the sample point is HHT, HTH or THH and X = 3 if the sample point is HHH.

We can make a distinction between a discrete random variable and a continuous variable. A random variable defined on a discrete sample space is known as a discrete random variable and it can assume either only a finite number or a countably infinite number of values. The number of car accident, the number of heads etc. are examples of discrete random variables.

A continuous random variable, like height, weight etc. is a random variable defined on a continuous sample space and assuming an uncountably infinite number of values.

The probability distribution of a random variable may be defined as a statement expressing the different values taken by a random variable and the corresponding probabilities. Then if a random variable X assumes n finite values X₁, X₂, X₃, …….., X _n with corresponding probabilities P₁, P₂, P₃, …….., P_n such that

(i) p_i ≥ 0 for every i …………… (15.33)
and (ii) ∑ p_i = 1 (over all i) ………….. (15.34)

then the probability distribution of the random variable X is given by

For example, if an unbiased coin is tossed three times and if X denotes the number of heads then, as we have already discussed, X is a random variable and its probability distribution is given by

There are cases when it is possible to express the probability (P) as a function of X. In case X is a discrete variable and if such a function f(X) really exists, then f(X) is known as Probability Mass Function (PMF) of X, f(X), then, must satisfy the conditions:

When x is a continuous random variable defined over an interval [ β , β ], where β> β , then x can assume an infinite number of values from its interval and instead of assigning individual probability to every mass point x, we assign probabilities to interval of values. Such a function of x, provided it exists, is known as probability density function (pdf) of x. f(x) satisfies the following conditions:

and the probability that x lies between two specified values a and b, where α ≤ a < b ≤ β , is given by

Expected value of a Random Variable

Expected value or Mathematical Expectation or Expectation of a random variable may be defined as the sum of products of the different values taken by the random variable and the corresponding probabilities . H en ce, i f a random vari able x assumes n val ues x₁, x₂, x₃ … ., x _n with corresponding probabilities p₁, p₂, p₃ …., p_n , where pi's satisy (15.33) and (15.34), then the expected value of x is given by

Expected value of x² in given by

In particular expected value of a monotonic function g (x) is given by

Variance of x, to be denoted by , σ² is given by

The positive square root of variance is known as standard deviation and is denoted by  .

If y = a + b x, for two random variables x and y and for a pair of constants a and b, then the mean i.e. expected value of y is given by

and the standard deviation of y is

When x is a discrete random variable with probability mass function f(x), then its expected value is given by

Where

For a continuous random variable x defined in [–∞, ∞], its expected value (i.e. mean) and variance are given by

Properties of Expected Values

Example 15.27: An unbiased coin is tossed three times. Find the expected value of the number of heads and also its standard deviation.

Solution: If x denotes the number of heads when an unbiased coin is tossed three times, then the probability distribution of x is given by

The expected value of x is given by