Unit 3: Probability

Introduction and Importance of Probability

Probability is one of the important branches of Statistics that is concerned with random phenomena.

Everyday concept of “likelihood”, “predictability” and “certainty” are formalized by the part of statistics called probability. Probability is a statistical method used to study of the chance of occurrence and non-occurrence of different phenomena. It is the synonym of the words most likely, chance, probably etc. Hence, Probability defined as „uncertainty is numerically expressed.

The theory of probability was developed in the middle of the seventeenth century. It was originated from the problem related to gambling such as tossing a coin, throwing a die, drawing cards from a pack of cards etc. Therefore, in the early days it was known as the theory of games and chance. But nowadays, there is hardly any displine (field) where “probability” has not been used. The theory of probability is used in almost all displine like physical science, natural science, biological science, medical science, engineering, social sciences, business, industry etc.

Definition:

Probability is a numerical measure (with a value lying between 0 and 1) of the likelihood or chance that a particular event will occur or not.

Probability is simply a number lies between zero and one. That is;. It is generally denoted by 'P'.

Basic Terminologies in Probability

Random Experiment

An experiment is called a random experiment if it performed a large number of times under essentially homogeneous conditions; the result is not unique but may be any one of the various possible outcomes. Tossing an unbiased coin and rolling a uniform die are some examples of random experiment.

Sample Space

The set of all possible outcomes of a random experiment is called sample space. Each outcome is thus considered as a sample point in the sample space. It is usually denoted by S. For example, in tossing an unbiased coin once, there are two possible outcomes head (H) and tail (T). So, the sample space is given by S =

Trial and Event

Performing a random experiment is called a trial and outcome or combination of outcomes (i.e. results) of random experiment is called an event. For example; flipping an unbiased coin is a trial and getting either head or tail is an event.

Exhaustive Cases

The total number of possible outcomes of a random experiment is called the exhaustive cases for the experiment. In a toss of single coin, we can get head (H) or tail (T). Hence, exhaustive number of cases is 2

Similarly, in rolling a six faced die, the exhaustive number of cases = 6

Favourable Cases or Events

The number of outcomes of a random experiment which result in the happening of an event are termed as the cases favourable to the event. In other words, the number of outcomes which are in favour of happening of an event are called favourable cases. For example; in drawing a card from a pack of 52 playing cards, the number of cases favourable to drawing a queen is 4. Similarly , in a toss of two coins, the number of cases favourable to the event „exactly one head „ is 2, viz., HT , TH and for „two heads‟ is one viz., HH.

Mutually Exclusive Events

Two or more events are said to be mutually exclusive if the happening of any one of them excludes the happening of all others in the same experiments i.e. if two or more than two events cannot occur simultaneously at the same time in the same trial. For example in a single tossing of a coin, we may get either head or tail but not both. Thus, the events head and tail in a single tossing of a coin are mutually exclusive. Similarly, in the throw of a die, the six faces numbered 1, 2, 3, 4, 5 & 6 are mutually exclusive. Thus, no two or more of them can happen simultaneously.

Independent Events

Events are said to be independent if the occurrence of one event does not affect the occurrence of the other events and vice versa. In tossing of a coin, the occurrence of head in first tossing is independent of the occurrence of head in the second tossing. Similarly, drawing of balls from an urn gives independent events if the draws are made with replacement.

Dependent Events

Events are said to be dependent if the occurrence of one event affects the occurrence of the other events and vice versa. For example, if a pack of cards is playing without replacemement, the occurrence of a king in first draw affects the occurrence of other cards in the second draw.

Fundamental Principle of Counting

The counting rules facilitate to calculate the all possible outcomes in the experiment.

The fundamental principle of counting states as follows:

If there are 'k' separate parts of an experiment, and the first part can be done in n₁ ways, second successive part in n₂ ways ....... and k^thsuccessive part in n_k ways, then the total number of possible outcomes is given by n₁ × n₂ × ... × n_k.

Permutation

The literal meaning of the word permutation is “Arrangement”. Therefore, permutation is the arrangement of objects taken some or all at time in some order. If there are 'n' objects and they are to be placed in any definite arrangement or order. The number of permutations of 'n' different objects taken 'r' objectives at a time is denoted by ⁿp_r or P (n, r) and defined as

where, n! = n (n – 1) (n – 2)! r n

Combination

The literal meaning of the word combination is “Selection”. Therefore, combination is the selection of objects taken some or all at a time without specific order. Note that the position of the object is meaningless in combination.

A combination of 'n' different objects taken 'r' objects at a time, is denoted by ⁿC_r or C (n, r) or and defined to be

ⁿC_r for r n

Approaches of Probability

The chance of occurrence or non-occurrence of any event in a random experiment is called probability. There are four approaches to define the probability.

— Mathematical or classical or priori approach

— Statistical or empirical or relative frequency approach

— Subjective approach

— Axiomatic approach

Mathematical or Classical or Priori, approach of Probability

Let n be the exhaustive, mutually exclusive and equally likely cases (or outcomes) out of which 'm' are favourable cases to the happening of an event A. Then the probability of happening (occurrence) of an event A is given by

P (A) =

or, P (A) = . . . . . . . .(1)

P (A) + P ( ̅) = 1

P ( ̅) = 1 – p(A)

Statistical or Empirical or Relative Frequency Approach

This approach of probability is based on statistical data. If an experiment is performed repeatedly a large number of times under essentially homogeneous and identical conditions, then the limiting value of the ratio of the number of times the event occurs to the total number of trials of the experiment as the number trials becomes indefinitely large, is called the probability of happening of the event. It being assumed that the limit is finite and unique.

Suppose an event A occurs m times in n repetitions of a random experiment. Then the ratio gives the relative frequency of the event A. In the limiting case when n becomes sufficiently large, then a number which is called the probability of A.

Symbolically,

P (A) =

Subjective Approach

The subjective probability approach is purely individualistic in nature. Therefore, this approach of probability is completely based on the personal beliefs, feelings, experience, judgement, personal discretion of a person. Since, different persons may assign different probabilities one cannot arrive at objective conclusions using probabilities assigned by this subjective method. This method of assigning probability is generally used by top level authorities on the basis of their discretion

Axiomatic Probability

The modern theory of probability is based on the axiomatic approach introduced by the Russian mathematician A.N. Kolmogorov in 1930. The axiomatic definition of probability includes both classical and empirical definition of probability and at the same time is the free from their drawbacks. It is also called elementary properties of probability. It states the following facts:

Definition: Given a sample space of a random experiment, the probability of occurrence of any event „E‟ is defined as a set function P( ) satisfying the following axioms:

Axiom 1: Axiom of non-negativity

P( ) is defined, is real and non-negative) i.e. P (E) ≥ 0

Axiom 2: Axiom of certainty

The sum of the probabilities is unity i.e. P , where S = Sample space

Axiom 3: Axiom of additivity

If is any finite or infinite sequence of disjoint events of sample space S, then

P or P

Note: When probability is not given directly, then it can be categorised into two cases:

Case (i): When one item is selected at a time.

Case (ii): When more than one item is selected at a time.

Examples related to

Case (i): When one item is selected at a time.

Probability of getting an event A, P (A) =

Where, n = Total number of cases m = Favourable number of cases Example 1

A card is drawn at random from a well-shuffled pack of 52 cards. What is the probability of getting?

(a) a red card,

(b) a black king Solution:

There are 52 cards in a pack. Total number of cases (n) = 52 (a) P (a red card) =?

Since, there are 26 red cards.

Favourable number of cases (m) = 26

Required probability of getting a red card is

P (a red card) =

(b) P (a black king) =?

There are 2 black kings

Favourable number of cases (m) = 2

Required probability of getting a black king is

P (a black king) =

Example 2

Twenty balls are numbered from 1 to 20. If one ball is drawn at random, what is the probability that the ball drawn is multiple of 4 or 7?

Solution:

Total number of cases (n) = 20

Favourable number of cases (m) = The number of cases which are multiple of 4 or 7

i.e. {4, 7, 8, 12, 14, 16, and 20}

i.e. m = 5+2 = 7

Required probability that the ball drawn is multiple of 4 or 7 =

Examples related to

Case (ii): When more than one item is selected at a time.

Probability of getting the event A, P(A) =

Where, n = Total number of cases = Selection of r out of total

m = Favourable number of cases

= Selection of r out of

Example 3

A bag contains 8 red, 4 white and 5 black coloured balls. Three balls are drawn randomly from a bag. Find the probability that (i) all are red (ii) 2 is red and 1 white (iii) 2 are red and 1 other (iv) all colour balls.

Solution:

Exhaustive (total) number of cases (n) = number of cases of selection of 3 balls out of 17 ball

= ¹⁷C₃ = 680

(i)Favourable number of cases of drawing all 3 are red balls (m) = ⁸C₃ = 56

P (all are red) =

(ii) Favourable cases for 2 red and 1 white (m) = ⁸C₂ × ⁴C₁

= 28× 4= 112

P (2 is red and 1 white) =

(iii)Favourable cases for 2 are red out of three drawn balls i.e. 2 are red and 1 other

(m) = ⁸C₂ × ⁹C₁ = 28×9 = 252

P (2 are red and 1 other) =

(iv) Favourable number of cases for all coloured balls

(m) = ⁸C₁ × ⁴C₁ × ⁵C₁ = 8 × 4 × 5 = 160

P (all colour balls) =

PU 2019(Spring), 2016(Spring)

2(a) From a group of 3 Nepalese, 4 Indians and 5 Americans, a sub –committee of 4 persons are selected at randomly. Find the probability that the sub-committee will consist (i) 3 Nepalese and 1 Indian (ii) 1 Nepalese , 1 Indian and 2 American (iii) 4 American.

Solution:

Total persons = 3 Nepalese + 4 Indians + 5 Americans =12

Total number of cases (n) = Selection of 4 persons out 12

= A sub-committee of 4 persons out of 12 = = 495

(i) P(a sub-committee will consist 3 Nepalese and 1 Indian)= ?

Favourable number of cases (m) = Number of cases of a sub committee will consist 3 Nepalese and 1 Indian

= = 1 = 4

P (a sub-committee will consist 3 Nepalese and 1 Indian)=

(ii) P (1 Nepalese, 1 Indian and 2 American) =?

Favourable number of cases (m) =

= 3 120

P (1 Nepalese, 1 Indian and 2 American) =

(iii) P (a sub-committee will consist all 4 American) =?

Favourable number of cases (m) =

= 1

P (a sub-committee will consist all 4 American)

PU2015 (Fall)

In a company, there are 4 civil engineers, 5 IT officers and 6 finance officers. A committee of 3 members has to be formed from random selection. What is the probability that committee consists of (i) All civil engineers

(ii) One civil engineer or one civil engineer & 2 others

(iii)At least one civil engineer Solution:

Total members = 4 + 5 + 6 = 15

Total number of cases (n) = Number of cases of selection of 3 members out of 15 members

= = 455

(i) P (All civil engineers) =?

Favourable number of cases (m)= selection of 3 civil engineers out of 4 civil engineers = = 4

P (All civil engineers) =

(ii) P (One civil engineer) =?

Favourable number of cases (m)= selection of 1 civil engineer out of 4 civil engineers and 2 members out 11 others = = 4

P (One civil engineer)

(iii) P ( At least one civil engineer ) = ?

Favourable number of cases (m) = Number of cases of no civil engineer = Selection of 3 members out of 11 non civil engineers = = 165

P( At least one civil engineer ) = 1 – P(No civil engineer )

= 0.637

Additional

(iv) P (each of one kind) =?

P (1 civil engineer, 1 IT officer & 1 finance) =?

Favourable number of cases (m) = = 4

P (each of one kind) =

PU 2013 (Fall)

In a organization, there are 2 civil engineers, 4 IT officers and 2 accounts. A committee of four members has to be formed. What is the probability that a committee contains

(i) one civil engineer and 3 IT officers ?

(ii) no civil engineer ?

(iii) At least one IT officers?

Solution:

Total members = 2+ 4 + 2 = 8

Since, a committee of four members has to be formed

Total number of cases (n) = Number of cases of selection of 4 members out of 8 members

= = 70

(i) P (one civil engineer and 3 IT officers) =?

Favourable number of cases (m) = selection of 1 civil engineer out of 2 civil engineers and 3 IT officers out of 4 IT officers = = 24= 8 P(one civil engineer and 3 IT officers) =

(ii) P(no civil engineer ) = ?

Favourable number of cases (m) = selection of no civil engineer = = 1

P(no civil engineer ) =

(iii) P(At least one IT officers) = ?

Favourable number of cases (m) = Number of cases of no IT officers = Selection of 4 members out of 4 non IT officers = = 1

P( At least one IT officers ) = 1 – P(No IT officers )

= 0.985

Example 4

Five men in a group of 20 are graduates. If 3 are chosen out of 20 at random, what is the probability that

a) all are graduates b) none of them is graduate

c) at least one of them being graduate

Solution: Here,

Total number of men = 20

Number of graduates = 5

Number of non-graduates = 20 – 5 = 15

If three are chosen at random

Total number of possible outcomes (n) = ²⁰C₃= 1140

(a) P (all are graduates) =?

Favourable number of cases (m) = ⁵C₃ = 10

P (all are graduate)

(b) P (none of them are graduate) =?

Favourable number of cases for none of them are graduate (m) = ¹⁵C₃ = 455

P (none of them are graduate) =

= 1 – P (none them are graduate)

= 0.60

Laws (or rules) of Probability)

Following are the laws of probability:

— Additive law of probability

— Multiplicative law of probability

Additive Law of Probability (Addition Theorem of Probability)

(At least one, one of them, or) “Union”

Case I: when events are not mutually exclusive

— If A and B are not mutually exclusive events, then the probability of occurrence of at least one of them is given by

P (A or B) = P (AB) = P (A) + P (B) – P (AB)

The probability of occurrence of at least one of them can also be written as

P (AB) = 1 – = 1 –

Demorgan’s law

— If A, B and C are three not mutually exclusive events then the probability of occurrence of at least any one of them is given by

P (A or B or C) = P (ABC) = P (A) + P (B) + P(C) – P (AB) – P (BC) – P (CA) + P (ABC)

Additive Law of Probability (Addition Theorem of Probability)

Case II: when events are mutually exclusive

Let A and B be two mutually exclusive events. Then the probability of the occurrence of either event A or event B is the sum of their individual probabilities. Hence, the probability of occurrence of either event A or event B is given by

P (A or B) = P (AB) = P (A) + P (B)

If A, B and C are three mutually exclusive events, then the probability of occurrence of either events A or B or C is given by

P (A or B or C) = P (ABC) = P (A) + P (B) + P(C)

Example 5

The probability that a company execute will travel by plane is and that he will travel by train is. Find the probability of his travelling by plane or train.

Solution: Let A and B be the events that a company execute will travel by plane and train respectively.

P (AB) = P (A) + P (B)

Multiplicative Law of Probability (or Multiplication Theorem of Probability)

(And, both, as well as, all, three, two) “Intersection”

Case (i): When events are independent

(a) If there is no change in total number (i.e. with replacement)

(b) If dependent condition is not stated i.e. conditional case is not given

Let A and B are two independent events, then the probability of occurrence of both the events is the product of their individual probabilities.

Symbolically,

P (A and B) = P (AB) = P (A) P (B)

If A, B and C are three independent events, then

P (ABC) = P (A) .P (B). P(C)

Multiplication Theorem of Probability

(And, both, as well, two, three, all, )

Case (ii): When the events are dependent:

(a) If there is change in total numbers (i.e. without replacement or not replaced , transferred)

(b) If dependent condition is stated i.e. conditional case is given

If A and B are two dependent events, then the probability of simultaneous happening of two events A and B is given by

Similarly,

Where, is the conditional Probability of the occurrence (happening) of event B given that (if) event A has already occurred (happened).

& is the conditional Probability of occurrence (happening) of event A given that event B has already occurred (happened).

If A, B and C are three dependent events, then

Where, is the conditional probability of occurrence (happening) of event C given that (if) both events A and B have already occurred (happened).

Conditional Probability

Definition: Conditional probability is the probability that an event will occur given that another event has already occurred. If A and B are two dependent events, then the conditional probability of event A given that (if) event B has already occurred is given by,

P (A/B) =, provided P (B) 0

Similarly, the conditional probability of event B given that (if) event A has already occurred is given by P (B/A) = , provided P (A) 0

PU 2014(Fall), 2017(Spring)

A problem in statistics is given to three students A, B, and C whose chances of solving it are and respectively. If they independently solve it. Find the probability that

(i) All can solve the problem.

(ii) None can solve the problem.

(iii)Problem can be solved. (or at least one of them will solve the problem) (iv) Exactly one can solve the problem

Only one of them can solve the problem.

(v) A solves it but B and C cannot.

Solution:

Probability that A solves a problem i.e. P (A) =

and P (

Probability that B solves a problem i.e. P (B) =

and P (

Probability that C solves a problem i.e. P(C) =

and P (

(i) P(All can solve the problem) = P (ABC)

= P (A) P (B)P(C)

= 0.25

(ii) P(None can solve the problem) = P ()

= P ( ̅) P ( ̅) P ( ̅)

(iii) P(problem can be solved)=

= P (A)+P (B)+P(C)–P (AB)–P (BC)–P(CA)+P ABC)

= P (A) + P (B) + P(C) – P (A) P (B) – P (B) P(C) – P(C) P (A)+ P (A) P (B) P(C)

= 0.95 are independent events.

Alternatively,

P(problem can be solved) =

= 1- P (

= 1 - P ()

= 1 – P ( ̅) P ( ̅) P ( ̅)

(iv) P (Exactly one can solve the problem)

= P (A) + P (B) + P(C)

= P (A) P ( ̅) P ( ̅) + P ( ̅) P (B) P ( ̅) + P ( ̅) P ( ̅) P(C)

(v) P (A can solve it but B and C cannot)

= P (A)

= 0.041

Example 6

A, B and C will pass a certain examination in the proportion 2:4:6. What is the probability that

(i) at least two of them will pass the examination?

(ii) at least one of the them will pass the examination?

(iii) all will pass the examination?

(iv) none will pass the examination?

(v) B will pass but not A and C?

Solution:

Since, the pass proportion of A, B and C in a certain examination are in the rato 2:4:6.

Total ratio = 2 + 4 + 6 = 12. Then

(i) P (at least two of them will pass the examination)

= P

= P P P P + P

( Events are mutually exclusive and independent)

(ii) P (at least one of the them will pass the examination) =

= 1– P

= 1 – P ( ̅) P ( ̅) P ( ̅)

= 1– 0.2777

= 0.7222

(iii) P

= P

(iv) P

= P

= P P P ( ̅)

= 0.2777

(v) P

P (

= P ( ̅) P ( ̅)

Example 7

A construction company is bidding for two contracts A and B. The probability that the company will get contract A is , the probability that the company will get contract B is and the probability that the company will get both the contracts is. What is the probability that the company will get contract A or B?

Solution:

We have,

The probability that the company will get contract A or B is given by

P (AB) = P (A) + P (B) – P (AB)

(i) P ( at least one machine will work without failure )

= 1– P

= 1 – P ( ̅) P ( ̅) P ( ̅)

= 1 – = 1– 0.28 = 0.72

(ii) P (at least two machine will work without failure)

= P

= P P P P + P

( Events are mutually exclusive and independent)

(iii) p (all machines will work without failure)

= P

= 0.03

Example 8

Two brothers: Mr. X and Mr. Y appear in an interview for getting the scholarship. The scholarship can be provided for two persons. The probability of getting scholarship by Mr. X is and getting by Mr. Y is . What is the probability that,

(a) both of them will get scholarship.

(b) Only one of them will get scholarship.

(c)None of them will get scholarship.

Solution:

Let P (A) = Probability that Mr. X will get scholarship

= Probability that Mr. X will not get scholarship P (B) = Probability that Mr. Y will get scholarship

= Probability that Mr. X will not get scholarship

Given that,

P (A) =

P (B) =

(a) Probability that both of them will get scholarship,

P (AB) = P (A) P (B)

Events are independent

(b)Probability that only one of them will get scholarship

= P (AB)

= P (A) + P (B)

= P (A). P ( ̅) + P ( ̅). P (B)

(c ) Probability that non of them will get scholarship

P () = P ( ̅) P ( ̅)

(b)Probability that only one of them will get scholarship

= P (AB)

= P (A) + P (B)

= P (A). P ( ̅) + P ( ̅). P (B)

(c ) Probability that none of them will get scholarship

P () = P ( ̅) P ( ̅)

(i)Required probability that both will be alive 25 years

P (AB) = P (A) × P (B) = 0.3 × 0.4 = 0.12

(ii) Probability that only the man will be alive

P (A) = P (A) × P ( ̅) = 0.3 × 0.6 = 0.18.

(iii) Probability that the only woman will be alive

= P (B) = P (B) × P) = 0.4 × 0.7 = 0.28

(iv) Probability that none will be alive

P ( ) = P) P ( ̅) = 0.7

(v) Required probability that at least one of them will be alive

= 1 - P ( )

= 1-

= 0.58

(vi) Probability that only one of them will be alive

= P (A or B) = P (A

= P (A) × P ( ̅) + P ( ̅) × P (B)

= 0.3 × 0.6 + 0.7 × 0.4

= 0.46

Example 9

If the probability of machines and working without failure are 0.2, 0.3 and 0.5 respectively. Find the probability that

(i) at least one machine will work without failure.

(ii) at least two machine will work without failure.

(iii) all machines will work without failure.

Solution:

Let A, B and C denote the events that machines and working without failure respectively.

Then

P, P

(i) P(at least one machine will work without failure )

= 1– P

= 1- P

= 1 –P()

= 1 – P ( ̅) P ( ̅) P ( ̅)

= 1 –

= 1– 0.28 = 0.72

(ii) P (at least two machine will work without failure)

= P

= P PP+ P

( Events are mutually exclusive and independent)

(iii) P (all machines will work without failure) = P

= P

= 0.03 PU 2018 (Fall)

Define axiomatic approach of probability. In certain locality of town, a survey of 600 women about the fuels for cooking revealed that 230 use kerosene oil, 175 women use electricity and 40 women use both. Find the probability that a women selected use;

(i) At least one of the fuel.

(ii) Neither of the fuel.

(iii) Kerosene oil only.

(iv) Electricity only (v) Exactly one.

Solution: Let K and E be the events that women in certain locality use kerosene oil and electricity for cooking respectively.

Total number of cases (n) = 600

P(K) = = 0.383, P( ̅) = 1-0.383 = 0.617

P( E) = = 0.292, P(̅ ) = 1-0.292 = 0.708

P(K

(i) P(At least one of the fuel ) = P(K

P(K P(K) + P( E) - P(K

= 0.383 + 0.292 – 0.067

= 0.608

(ii) P (Neither of the fuel) = P (

P ( ) = 1 - P(K

= 1 – 0.608

= 0.392

(iii) P (Kerosene oil only) = (K) =?

(K) = P (K) - P (K

= 0.383 - 0.067

= 0.31

(iv) P(Electricity only) = (E) = ?

(E) = P(E) - P(K

= 0.292 - 0.067

= 0.225

(V) P(Exactly one ) = (K) + (E)

= 0.31 +0.225

= 0.535

Example 10 The probability that a manufacturer will produce „brand X‟ product is 0.13, the probability that he will produce „brand Y‟ product is 0.28 and the probability that he will produce both brand is 0.06. What is the probability that the manufacturer who has produced „brand Y‟ will also have produced „brand X‟?

Solution: Let X and Y be the events that a manufacturer will produce brand X and brand Y respectively.

P(X) = 0.13, P(Y) = 0.28, P (,

The probability that the manufacturer who has produced „brand Y‟ will also have produced „brand X‟ is

Example 11

In a certain school, 20% students failed in English, 15% students failed in Mathematics and 10% students failed in both English and Mathematics. A student is selected at random. If he failed in English, what is the probability that he also failed in Mathematics.

Solution:

Let E and M be the events that denote the students failed in English and failed in Mathematics respectively. Then

P (E) = 20% = 0.2

P (M) = 15% = 0.15

P (EM) = 10% = 0.10

If one student is selected at random, the Probability that if he failed in English then he also failed in Mathematics is given by

P (M/E) =

PU 2016 (Fall)

Q.No.2 (a)

An oil exploration company currently has two active projects, one in Asia and the other in Europe. Let A be the event that the Asian project is successful and B be the event that the European project is successful. Suppose that A and B are independent events with

P (A) = 0.4 and P (B) = 0.7

(i)If the Asian project is not successful, what is the probability that the European project is also not successful? Explain your reasoning.

(ii) What is the probability that at least one of the two projects will be successful?

(iii)Given that at least one of the two projects is successful, what is the probability that only the Asian project is successful?

Solution:

P (A) = 0.4, P ( ̅ ) = 1 - P (A) = 1- 0.4 = 0.6

P (B) = 0.7, P ( ̅ ) = 1 - P (B) = 1- 0.7 = 0.3

(i) If the Asian project is not successful, the probability that the European project is also not successful is given by

A and B are independent events

= P (AB) = P (A) P (B)

P ( ̅ / ̅ ) = 0.3

P ( ̅ / ̅ ) =

This result concludes that if A and B are two independent events then conditional probability of event ̅ given ̅ is equal to unconditional probability of event i.e. P( ̅ / ̅ ) =P( ̅ )

(ii) The probability that at least one of the two projects will be successful is given by

P (A P(A) + P( B) - P(A

= P(A) + P( B) - P( A) P( B)

= 0.4 + 0.7 – 0.4

= 0.82 A and B are independent events

P (AB) = P (A) P (B)

(iii) Given that at least one of the two projects is successful , the probability that only the Asian project is successful is given by

P (A/A

PU 2015 (Spring)

Q.N0. 2(b)

The HAL Corporation wishes to improve the resistance of its personal computer to disk driver and key board failures. At present, the design of computer is such that disk –drive failures occurs only one third as often as keyboard failure. The probability of simultaneous disk-drive and key board failure is 0.05.

(i) If the computer is 80% resistant to disk and / or keyboard failure, how must the disk-driver failure probability be?

(ii) If the keyboard is improved so that it fails only twice as often as the disk-driver (and the simultaneous failure probability is still 0.05), will the disk drive failure probability from part (i) yield a resistance to disk-drive and/or keyboard failure higher or lower than 90%

Solution:

Let D and K be the events that the computer has disk-driver failure and key-board failure respectively.

By given condition

The disk –drive failures occurs only one third as often as keyboard failure

P(D) = P(K)

P (K) = 3 P(D)

& The probability of simultaneous disk-drive and key board failure is 0.05.

i.e. P(D

(a) Since, the computer is 80% resistant to disk and / or keyboard failure

P ( ) = 0.80

P(D

= 1 – 0.80

= 0.20

P (D) =?

Now,

P (D P (D) + P (K) - P (D

or, 0.20 = P(D) + 3 P(D)- 0.05 or, 0.25 = 4 P(D)

or,

P(D) = 0.0625

The probability of disk driver failure is 0.0625

(ii) P(D) = 0.0625

P(K) = 2 P(D)= 2 = 0.125

P(D

P(D P(D) + P( K) - P(D

= 0.0625 + 0.125 -0.05

= 0.1375

P(D=0.1375

Hence, the probability of resistance to disk drive and/or keyboard failure is

P( ) = 1- P(D

= 1 -0.1375

= 0.8625

= 86.25% which is lower than 90%

Applications of Conditional Probability

Bayes's Theorem (Rule for the Inverse Probability)

Statement: Let be n mutually exclusive (disjoint) and exhaustive events of sample space S with probabilities P for i= 1, 2, 3,. . . .,n. For any arbitrary event A which is a subset of the given sample space S such that, then the probability that it was preceded by the particular event is given by

, i=1,2,3, . . . ,n

Proof:

Let S be the sample space and it can be partioned (divided) into various mutually exclusive (disjoint) subsets with P for i= 1, 2, 3,. . . .,n . Let A be any arbitrary event which is subset of sample space S and having some intersecting part with mutually exclusive and exhaustive events .

Then from figure

A =

P(A)=P

Since, , , . . . . . .. , are mutually exclusive (disjoint) events. Then, by addition theorem of probability

P(A)= P+ P+P

P(A) = . . . . . . .(i)

For any particular event , the conditional probability P is given by

P = P (A) P

or, P

, i= 1, 2, 3,. . . .,n [From (i)]

Where,

P (E₁), P(E₂) ...., P(E_n) which are already known before conducting an experiment is termed as priori probabilities.

The conditional probabilities P (E₁/A), P (E₂/A),. . . . . .,P (E_n/A) which are computed after conducting the experiment are termed as posteriori probabilities

Example 12

Assuming that a factory has two machines and Past record showed that machine produces 30% of the items of output and machine produces 70% of the items. Further, 5% and 1% of the items produced by machine and respectively were defective. If a defective item is selected at random what is the probability that it was produced by machine ?

Solution: Let and be the events of producing items of output by machines and respectively.

= 0.30

= 0.70

Let A be the event of producing defective item. Then,

= 0.05

= 0.01

= ?

By Bayes‟s theorem

The required probability that the defective item was produced by machine is

Example 13

A company buys tyres from two suppliers, 1 and 2. Supplier 1 has a record of delivering tyres containing 10% of the defectives, whereas supplier 2 has a defective rate of only 5%. Suppose 40% of the current supply came from supplier 1. If a tyre is taken from this supply and observed to defective, find the probability that it is from a supplier 1.

Solution:

Let and be the events of delivering tyres from suppliers 1 and 2 respectively.

= 0.40

=1-0.40 = 0.60

Let A be the event of delivering defective tyres. Then,

= 0.10 = 0.05

= ?

By Bayes‟s theorem

0.571429

The required probability that the defective tyre is from a supplier 1 is 0.571429

PU2019 (Spring)

2. b) In a certain factory machines I, II, III are all producing springs of the same length of their production machines I, II, III produces 2%, 1% and 3% defective springs respectively of the total production of springs in the factory, I machine produces 35%, machines II produces 25% and machine III produces 40%. If one spring is selected at random from total springs in a day, find:

i. the probability that it is defective ii. the conditional probability that it was produced by machine III.

Solution:

Let and be the events of producing springs of the same length by machines I, II and III respectively. Then

= 0.35

= 0.25

= 0.40

Let A be the event of producing defective springs by three machines. Then,

= 0.02 = 0.01

= 0.03

(i) Probability that defective springs produced by machines is

= 0.35

= 0.0215

(ii) By Bayes‟s theorem

The probability that defective spring was produced by machine III is given by

PU: 2018(Spring)

2. a) A firm has 80% of its service calls made by a contractor and 10% of these calls result in customer complaints. The other 20% of the service calls are made by their own employees, and these calls have 5% complaint rate. Find the probability of a complaint. Also, using Bayes theorem to find the probability that a complaint was from a customer whose service was provided by the contractor.

Solution: Let and be the events that a firm‟s service calls made by a contractor and by their own employees respectively.

= 0.80

= 0.20

Let A be the event that a complaint from customer. Then

= 0.10

= 0.05

The probability of a complaint is given by

= 0.80

= 0.09

By Baye‟s theorem

The probability that a complaint was from a Customer whose service was provided by the contractor is given by

PU: 2017 (Fall), 2013(Spring)

2. a) State Bayes theorem. In a class of 75 students, 15 students were considered to be very intelligent, 45 as medium and rest below the average. The probability that a very intelligent student fails in examination is 0.005, the medium students failing has probability 0.05 and corresponding probability of a below average is 0.15. If a student is known to have passed the examination, what is the probability that he is below the average?

Solution: Statement: Let be n mutually exclusive (disjoint) and exhaustive events of sample space S with probabilities P for i= 1, 2, 3,. . . .,n. For any arbitrary event A which is a subset of the given sample space S such that, then the probability that it is was preceded by the particular event is given by

, i=1,2,3, . . . ,n

Let and be the events that denote for very intelligent, medium and below the average students respectively.

Let A be the event that a student have passed the examination. Then,

= 1- 0.005 = 0.995

= 1- 0.05 = 0.95

=1- 0.15 = 0.85

If a student is known to have passed the examination, the probability that he is below the average is given by

Where,

= 0.20

= 0.939

PU 2018 (Fall)

2 (a) The contents of urns I, II and III are as follows: 1 white, 2 black and 3 red balls ; 2 white balls, 1 black and 1 red balls and 4 white , 5 black and 3 red balls. One urn is chosen at random and two balls are drawn. They happen to be white and red. What is the probability that they come from urns I, II or III?

Solution:

Let and be the events of selection of urns I, II and III respectively.

Let A be the event of drawing a white ball and a red ball. Then

By Bayes‟s theorem

The required probabilities that white and red balls come from urns I, II or III respectively are given by

Where,

= 0.238

PU 2016 (Spring)

2(c) A manufacturing firm produces steel pipes in three plants with daily production volumes of 500, 1000 and 2000 units respectively. According to the past experience, it is known that the fraction of defective outputs produced by the three plants are respectively 0.005, 0.008 and 0.010. If pipe is selected from a day‟s total production and found to be defective. From which plant the defective pipe is expected to have been produced?

Solution: Let and be the events producing steel pipes by three plants I, II and III respectively. Then

Let A be the event that the defective steel pipes produced by three plants. Then,

= 0.005

= 0.008

= 0.010

= ?= ?, = ?

By Bayes‟s theorem, the required probabilities that the defective steel pipes are produced from plants

I, II and III respectively are given by

The probability that the plants produce defective pipe is

= 0.00871

> > therefore plant III produced more defective pipes than plants

I , II.

Example 14

The chances of X, Y and Z becoming manager will be of a certain company are 4:2:3. The probabilities that bonus scheme will be introduced if X, Y and Z become manager are 0.3, 0.5 and 0.8 respectively.

(i) What is the probability that bonus scheme will be introduced?

(ii)If the bonus scheme has been introduced, what is the probability that X is appointed as the manager? Solution:

Let and be the events of X, Y and Z becoming managers respectively. Then Total ratio = 4+2+3 = 9

Let A be the event of introducing bonus scheme. Then,

= 0.3

0.5

0.8

= ?= ?, = ?

The probability that the bonus scheme will be introduced is given by

By Bayes‟s theorem,

If bonus scheme has been introduced, then the probability that X is appointed as the manager is given by

PU: 2013 (Fall)

2. b) A company has two plants to manufacture the screws. Out of 10000 screws, plant I manufactures 80% of the screws and plant II manufactures 20%. At plant I, 85% screws are rated standard quality. At plant II, only 65% screws are rated standard quality.

i. One screw is selected at random from the whole consignment and was found standard quality, What is the probability that the selected screw was manufactured by plant I?

ii. What is the probability that the selected screw being defective?

Solution:

Let be the events of manufacturing screws by plant I and plant II respectively.

=0.80

= 0.20

Let A be the event that screws are rated standard quality. Then,

= 0.85

0.65

(i) If screws are rated standard quality, the probability that the selected screw was manufactured by plant I is given by

Where,

= 0.80

= 0.81

(ii) The probability that the selected screw being defective

= 1- 0.839

= 0.161

PU 2015 (Spring)

2. a) Two production plants A and B, make wire cables that are sent to a common distributor. 40% of the cables sent to the distributor come from plant A, and the remaining 60% from plant B. Among the cables produced at Plant A, 95% meet the strength specifications; among the cables produced at plant B, 98% meet the strength specification.

i. The distributor selects one cable at random from among all cables in stock. If the cable selected is found to meet the strength specifications, what is the probability that the cable was produced at plant A?

ii. Let A be the event that a cable is manufactured at plant A; similarly, B be the event that a cable is manufactured at plant B. Let Y be the event that a cable meets the strength specifications; and let N be the event that a cable does not meet the strength specifications. (i) Are events A and Y independent? Give reasons.

Solution:

Let be the events of producing wire cables by plant A and plant B respectively.

=0.40

= 0.60

Let A be the event that the cables produced meet the strength specifications.

= 0.95

= 0.98

(i) If the cable selected is found to meet the strength specifications, the probability that the cable was produced at plant A is given by

P(A) =

(ii)

A = the event that a cable is manufactured at plant A

B = the event that a cable is manufactured at plant B

Y = the event that a cable meets the strength specifications

N = the event that a cable does not meet the strength specifications.

P (A) = 0.40

P(Y) = 0.968

P (A) P(Y) = 0.400.968 = 0.3872

P(Y P(A)P(Y/A)

= 0.40

= 0.38

Since, P(Y P(A) P(Y)

Hence, events A and Y are not independent

PU 2016 (Fall)

2. b) Three road construction firms, X, Y and Z, bid for a certain contract. From past experience, it is estimated that the probability that X will be awarded the contract is 0.40, while for Y and Z the probabilities are 0.35 and 0.25. If X does receive the contract, the probability that the work will be satisfactorily completed on time is 0.75. For Y and Z these probabilities are 0.80 and 0.70.

i. What is the probability that the work will be completed satisfactorily?

ii. It turns out that the work was done satisfactorily, what is the probability that Y was awarded the contract?

Solution:

Let and be the events of firms X, Y and Z awarded the contract respectively. Then