Probability

Plus Two

Dr.V.S.RaveendraNath M.Sc., M.Ed., Ph.D.

04/07/2023VSR 2

You've probably heard people say things like:

Probability - Introduction

Teen mother

The chance of rain tomorrow is 75%.

Teen mothers who live with their parents

He won the lottery!

All of these statements are about probability. We see words like "chance", "less likely", "probably" since we don't know for sure something will happen, but we realise there is a very good chance that it will.

04/07/2023VSR 3

Probability is a concept which numerically measures the degree of uncertainty and therefore is certainty of the occurrence of events.

Prologue Probability had its origin in the 16th century when an Italian physician and mathematician Jerome Cardon (1501 -1576) wrote the first book on the subject. “Book of Games of Chance”. Subsequently, the theory of probability was developed by Bernoulli, De-Moivar, Fisher and others.

04/07/2023VSR 4

The mathematicians is basically concerned

with drawing conclusions (or inference)

from experiments involving uncertainties.

For these conclusions and inferences to be

reasonably accurate, an understanding of

probability theory is essential.

In this section, we shall develop the

concept of probability with equally likely

outcomes

Introduction to Probability Theory

04/07/2023VSR 5

Experiment, Sample Space and Event

Experiment

This is any process of

observation or procedure that

can be repeated

(theoretically) an infinite

number of times and

has a well-defined set of

possible outcomes.

04/07/2023VSR 6

Sample spaceThis is the set of all possible outcomes of an experiment.

Event

This is a subset of the sample space of an experiment

Consider the following illustrations:

The set of all event is the power set of S , denoted By 2s .

04/07/2023VSR 7

Experiment 1: Tossing a coin.

Sample space: S = {Head or Tail}

Experiment 2: Tossing a coin twice.

S = {HH, TT, HT, TH}

Experiment 3: Throwing a die.

S = {1, 2, 3, 4, 5, 6}

04/07/2023VSR 8

Experiment 4: Two items are picked, one

at a time, at random from a

manufacturing process, and each item is

inspected and classified as defective or

non-defective.S = {NN, ND, DN, DD} whereN = Non-defectiveD = Defective

Some events:E1 = {only one item is defective} = {ND, DN}E2 = {Both are non-defective} = {NN}

04/07/2023VSR 9

Probability of an Event

Definition of a ProbabilitySuppose an event A can happen in m ways out of a total of n possible equally likely ways.Then the probability of occurrence of the event (called its success) is denoted by

n

m

Sn

AnAP

)(

)()( →no. of favourable

cases→no. of total outcomes of the experiment

04/07/2023VSR 10

The probability of non-occurrence of the event (called its failure) is denoted by

n

m

n

mnEP

1)(

Notice the bar above the E, indicating the event does not occur.

Thus, = 1

In words, this means that the sum of the probabilities in any experiment is 1.

04/07/2023VSR 11

* Odds in favour of A )(

)(

AP

AP

* Odds in against A )(

)(

AP

AP

Addition theorem for two or more events.i.e., P(A or B) = P(A) + P(B) – P(A B) . If A and B are mutually exclusive events then ,P(A or B) = P(A) + P(B) .

04/07/2023VSR 12

* In this chapter, we shall discuses the concept of Conditional probability of event. Baye's theorem, Multiplication rule of probability and random variable and its probability, Binomial distribution etc.

Conditional Probability

If E and F are two events associated with the same sample space of a random experiment, E given that F has occurred, is given by

0)(,)(

)()(

FP

FP

FEPEorFP

04/07/2023VSR 13

• Properties of conditional probability.

i) P(S or F) = P(F/F) = 1.

ii) If A and B are any two events of a sample space S and F is an event of S such that P(F) ≠ 0 then, P[(A B) or F] = P(A or F) + P(B or F) – P(A B) or F)

)(1) EorFPF

EPiii

04/07/2023VSR 14

* Multiplication Theorem on Probability.

P(E F) = P(E).P(F/E)=

P(F).P(E/F) .

Provided P(E) and P(F) ≠ 0 # More than two events E, F and G, then by multiplication rule of probability P(E F G) = P(E).P(F/E).P(G/EF). Where EF = E F.

04/07/2023VSR 15

Types of Event.

The given events are said to be Equally Likely, if none of them is expected to occur in preference to the other.

Two events are said to be Independent, if the occurrence of one does not depend upon the other.

Hence events E and F are independent event if P(EF) = P(E). P(F).

04/07/2023VSR 16

# For two independent events E and F,

the addition theorem becomes,

P(E or F) = P(E) + P(F) – P(E).P(F)

Or P(E or F) ).().(1)(1 FPEPFEP

04/07/2023VSR 17

Two Events

Let's consider "E1 and E2" as the event

that "both E1 and E2 occur".

If E1 and E2 are dependent events,

then:

P (E1 and E2) = P (E1) × P (E2 | E1)

If E1 and E2 are independent events,

then:

P (E1 and E2) = P (E1) × P (E2)

04/07/2023VSR 18

Three Events

For three dependent events E1, E2, E3, we

have

P(E1 and E2 and E3)

= P(E1) × P(E2 | E1) × P(E3 | E1

and E2)

For three independent events E1, E2, E3,

we have

P(E1 and E2 and E3) = P(E1) × P(E2) × P(E3)

04/07/2023VSR 19

Mutually Exclusive Events

Two or more events are said to be mutually exclusive if the occurrence of any one of them means the others will not occur (That is, we cannot have 2 events occurring at the same time).Thus if E1 and E2 are mutually exclusive events, thenP(E1 and E2) = 0.

04/07/2023VSR 20

Suppose "E1 or E2" denotes the event that "either E1 or E2 both occur", then(a) If E1 and E2 are not mutually exclusive events:P(E1 or E2) = P(E1) + P(E2) − P(E1 and E2)We can also write:P(E1 ∪ E2) = P(E1) + P(E2) − P(E1 ∩ E2)A diagram for this situation is as follows. We see that there is some overlap between the events E1 and E2. The probability of that overlap portion is P(E1 ∩ E2).

04/07/2023VSR 21

(b) If E1 and E2 are mutually exclusive

events:

P(E1 or E2) = P(E1) + P(E2)

Our diagram for mutually exclusive

events shows that there is no overlap:

04/07/2023VSR 22

Exhaustive event.A set of events is said to be exhaustive

if the performance of the experiment results in the occurrence of at least one of them. If a set of events E1, E2,……En , then for exhaustive events P(E1 E2 ….. En ) = 1.

If E1, E2,……En are mutually exclusive and exhaustive events and any events E is said to be compound events, if

0)(,).(

()(

1

1

n

ii

ii

n

ii

EifPE

EPEP

EEPEP

04/07/2023VSR 23

Partition of a sample space.

• The events E1, E2,……En represent a partition of the sample space S if they are pair wise disjoint, exhaustive and have non-zero probabilities.

a) Ei Ej = , i ≠ j , i, j = 1, 2, 3, ……..n.

b) Ei E2 …….. En = S

c) P( Ei ) > 0 for all i = 1, 2, ………n.

04/07/2023VSR 24

Theorem of Total Probability

P(A) = P(E1).P(A/E1) + P(E2).P(A/E2) +

…….+ P(En).P(A/En)

04/07/2023VSR 25

Bayes' Theorem

Let E1, E2 ......En are n non empty events

which constitute a partition of sample

space S, then

n

jjj

ii

EAPEP

EAPEPAEP

1

1

)/().((

)/().()/(

for any i = 1, 2, 3, ………n.

04/07/2023VSR 26

Binomial Distribution

P(r) , the probability of r successes, is given by:

rprnqCrnrP )(Where p = the probability of successes.q = the probability of failure.

\p + q = 1 , i.e., q = 1 – p .

Here n, p, q are called the parameters of Binomial Distribution.

04/07/2023VSR 27

Mean of binomial distribution

Mean = np

Variance V(X) = E(X2) – (E(X))2 .

Mean of probability distribution = XP (X)

Variance = X2P(X) – (mean)2

S.D. = X2P(X) – (mean)2

04/07/2023VSR 28

Variance of binomial distribution

Variance = npq

Standard Deviation ()

npqDS ).(. Recurrence Formula

).(..)1(

)()1( rP

q

p

r

rnrP

04/07/2023VSR 29

Mean of Random Variable:

n

inni PxPxpxPixxE

12211 .......)(

x1,x2 ……xn are possible values of the random

variable x with probabilities P1 , P2 …..Pn

respectively.

04/07/2023VSR 30

Asked Questions

Dr.V.S.RaveendraNath M.Sc., M.Ed., Ph.D.

Frequently

04/07/2023VSR 31

Q.1.A fair die is thrown, what is the probability that either an odd number or a number greater than 4 will up. (K.U)

Answer

Let S be the sample space throwing of the die S = {1, 2, 3, 4, 5, 6}. i.e., n(S) = 6Let A be the event of getting an odd number B be the number greater than 4 . \A = {1, 3, 5} ; B = {5, 6} ; also AB = {5} n(A) = 3; n(B) = 2 ; n(AB) = 1

6

1

)(

)()(

3

1

6

2

)(

)()(

2

1

6

3

)(

)()(

Sn

BAnBAP

Sn

BnBP

Bn

AnAP

)()()()( BAPBPAPBAP

3

2

6

123

6

1

3

1

2

1

The required probability = 2/3.

04/07/2023VSR 32

Q.2. A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a 6. Find the probability that it is actually 6. (AICBSE)

Answer

Let A be the man report that it is a 6. Let E1 be the event “6 has occurred” and E2 be that “6 has not occurred”

P(E1 ) = 1/6; P(E1 ) = 1–P(E1 ) = 1–1/6 =

5/6

also P(A/ E1 ) = ¾ . [ i.e., speak 3 out

of 4]

P(A/ E2 ) = 1-P(A/ E1 ) = 1-3/4 = ¼

04/07/2023VSR 33

)/().()/().(

)/().()/(

2211

111 EAPEPEAPEP

EAPEPAEP

Formulae

8

3

8

24

8

124881

255

8181

41.65

43.61

43.61

04/07/2023VSR 34

Q3. An insurance company insured 2000 Scooter drivers, 3000 Car drivers and 4000 Truck drivers. The probabilities of their meeting with an accident respectively 0.04, 0.06 and 0.15. One of the insured persons meets with an accident, find the probability that he is a Car driver. (AICBSE)

Answer

Let E1 , E2 , E3 , be the Scooter, Car and Truck drivers respectively.

Let A be person that meets with an accident

04/07/2023VSR 35

9

4

400030002000

4000)(

9

3

400030002000

3000)(

3

2

400030002000

2000)(

3

2

1

EP

EP

EP

100

1515.0)/(

;100

606.0)/(;

100

404.0)/(

3

21

EAP

EAPEAP

04/07/2023VSR 36

Required probability

)3/().3()2/().2()1/().1(

)2/().2()/2( EAPEPEAPEPEAPEP

EAPEPAEP

43

9

60188

18

10015.94

1006

.93

1004

.92

1006

.93

04/07/2023VSR 37

Q.4. Gowrave and Sowrave appear for an interview for two vacancies. The probability of Gowrave’s selection is 1/3 and that of Sowrave’s selection is 1/5. Find the probability that only one of them is selected. (CBSE)

Answer

Let A be Gowrave’s selection B be the Sowrave’s selection.

P(A) = 1/3 ; P(B) = 1/5.

)(AP Probability that Gowrave not selected = 1 – P(A) = 1 – 1/3 = 2/3.

04/07/2023VSR 38

)(BP Probability that Sowrave not selected

= 1 – P(B) = 1 – 1/5 = 4/5

Probability that only one of them is selected

5

215

2

15

4

5

1.3

2

5

4.3

1

)().()().(

BPAPBPAP

04/07/2023VSR 39

Q.5. In a factory which manufactures bolts, machines A, B and C manufacture respectively 25%, 35% and 40% of the bolts of their output 5, 4 and 2 percent are respectively defective bolts. A bolt is drawn at random from the total production and if found to be defective. Find the probability that it is manufactured by the machine B. (CBSE)

Answer

Let E1 , E2 and E3 be the events of manufacturing the bolt by machine A, B and C respective.

100

40)(;

100

35)(;

100

25)( 321 EPEPEP

04/07/2023VSR 40

Let A be the event that bolt drawn is defective

100

2)/(;

100

4)/(;

100

5)/(.,. 321 EAPEAPEAPei

Required probability

69

28

345

140

10040.

1002

.10035.

1004

10025.

1005

10035

1004

)()./()()./()()./(

)()./()/(

332211

222

EPEAPEPEAPEPEAP

EPEAPAEP

04/07/2023VSR 41

Q.6. A four digit number is formed using the digits 1, 2, 3, 5 with no repetitions. Find the probability that the number is divisible by 5 . (CBSE)

Answer

Number of favourable cases m = four digit numbers, which are divisible by 5 = 6.

The no. of exhaustive cases n = 4C4 = 4! =24.

Required probability = m/n = 6/24 = ¼

04/07/2023VSR 42

Q.7. A lot contains 50 defective and 50 non-defective bulbs. Two bulbs are drawn at random, one by one, with replacement. The events A, B and C are defined as the first bulbs is defective, the second bulbs is non-defective, the two bulbs are both defective or non-defective, respectively. Determine whether A, B and C are pair wise independent. Answe

r

Sample space S = {DD, DN, ND, NN} .Where D = Defective bulb and N = Non-defective bulb

04/07/2023VSR 43

Thus A = {DD, DN} , B = {DN, NN} , C = {DD, NN}.

P(A) = 2/4 = ½ , P(B) = 2/4 = ½ , P(C) = 2/4 = ½ .

P(A B) = P(DN) = ¼ ; P(B C) = P(NN) = ¼ P(A C) = P(DD) = ¼

Now P(A). P(B) = ½ ½ = ¼ = P(A B) i.e., A and B are independent Similarly B and C and also A and C are independentHence A, B, C are pair wise independent.

04/07/2023VSR 44

Q.8. An urn contains m white and n black balls. A ball is drawn at random and is put back into the urn along with k additional balls of the same colour as that of the ball drawn. A ball is again drawn at random. What is the probability that the ball drawn now is white. Answe

rLet W1 be : Ball drawn is white in the first draw. W2 be : Ball drawn is white in the second draw. and B1 be : Ball drawn is black in the first draw.

Now, P(W2) = P(W1) . P(W2 / W1) + P(B1) P(W2 / B1)

04/07/2023VSR 45

nm

m

knmnm

knmm

knmnm

mnmkm

knm

km

nm

m

knm

km

nm

m

))((

)(

))((

..

2

04/07/2023VSR 46

Q.9. A card is drawn out from a well shuffled pack of 52 cards. If E is the event “the card drawn out is a king or queen” and F is the event “the card drawn out is a queen or an ace”, find the probability P(E/F).

(CBSE)

Answer

P(F) = P( card is queen or ace) = 8/52 = 2/13.

P(EF) = P(card is queen) = 4/52 = 1/13

2

1

132131

)(

)()/(

FP

FEPFEP

04/07/2023VSR 47

Q.10. If P(A) = 3/5 and P(B) = 1/5 find P(AB) if A and B are independent events. Answe

rGiven that A and B are independent events. P(AB) = P(A).P(B)

= (3/5) (1/5) = 3/25

04/07/2023VSR 48

Q.11. If P(A) = 1/2 and P(B) = p ,P(AB) = 3/5 if A and B are mutually exclusive events.

Answer

Given that A and B are mutually exclusive events.

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

i.e., 3/5= ½ + p p = 3/5 – ½ = 1/10.

04/07/2023VSR 49

Q.12. A husband and his wife appear for an interview for two posts. The probability of husband’s selection is 1/7 and that of wife’s selection is 1/5 . What is the probability that only one of them is selected.?

Answer

Let A and B be the events of husband and wife respectively.

Here P(A) = 1/7 and P(B) = 1/5

5

4

5

11)(

7

6

7

11)(

BP

andA

04/07/2023VSR 50

Here A and B are independent events

7

2

35

1035

6

35

45

1

7

6

5

4

7

1

)().()().(.Re

BPAPBPAPyprobabilitqd

04/07/2023VSR 51

Q.13. A problem in Mathematics is given to three students whose chances of solving it are ½, 1/3, ¼. What is the probability in the following cases ? 1) That the problem is solved.

(Kerala)

2) only one of them solved correctly

(CBSE) 3) at teast one of them may solve it.

Answer

Let A, B, C be the three event when the problem in maths is solved by the three students.

04/07/2023VSR 52

4

3

4

11)(;

3

2

3

11)(;

2

1

2

11)(

.4

1)(;

3

1)(;

2

1)(

CPBPAP

CPBPAP

1) The probability that the problem is solved = Probability that the problem is solved by at least

one student.

4

3

4

11

4

3

3

2

2

11

)().().(1

CPBPAP

04/07/2023VSR 53

2) The probability that only one solves it correctly

)().().()().().()().().( CPBPAPCPBPAPCPBPAP

24

11

12

1

8

1

4

14

1

3

2

2

1

4

3

3

1

2

1

4

3

3

2

2

1

04/07/2023VSR 54

3) The probability that atleast one of them may solve the problem

4

3

4

11

4

3

3

2

2

11

)().().(1

CPBPAP

04/07/2023VSR 55

Q.14. In a bolt factory, machines A, B

and C manufacture respectively 25%,

35%, 40% of the total. Of their output 5,

4 and 2% are defective. A bolt is drawn

at random from the product.

1). What is the probability that the bolt

drawn is defective?

2) If the bolt drawn is found to be

defective, find the probability that it is a

product of machine B?.

04/07/2023VSR 56

Answer

100

2)/(;

100

4)/(;

100

5)/(

100

40)(;

100

35)(;

100

25)(

CDPBDPADandP

CPBPAHereP

Where D denotes defective bolt.

0345.0100

2.

100

40

100

4.

100

35.

100

5.

100

25

)/()()/()()/()()().1

CDPCPBDPBPADPAPDP

04/07/2023VSR 57

2) By Baye’s theorem

69

28

345

1401002

.10040

1004

.10035

1005

.10025

1004

.10035

)/().()/().()/().(

)/().()/(

CDPCPBDPBPADPAP

BDPBPDBP

04/07/2023VSR 58

Q.15. If the probability that person A will be alive in 20 years is 0.7 and the probability that person B will be alive in 20 years is 0.5, what is the probability that they will both be alive in 20 years? Answe

r

These are independent events, so

P(E1 and E2) = P(E1) × P(E2) = 0.7 × 0.5 = 0.35

04/07/2023VSR 59

Q.16. A fair die is tossed twice. Find the probability of getting a 4 or 5 on the first toss and a 1, 2, or 3 in the second toss. Answe

rP(E1) = P(4 or 5) = 2/6 = 1/3

P(E2) = P(1, 2 or 3) = 3/6 = 1/2

They are independent events, so

P(E1 and E2) = P(E1) × P(E2) = 1/3 × 1/2 = 1/6

04/07/2023VSR 60

Q.17. It is known that the probability of obtaining zero defectives in a sample of 40 items is 0.34 whilst the probability of obtaining 1 defective item in the sample is 0.46. What is the probability of(a) obtaining not more than 1 defective item in a sample?(b) obtaining more than 1 defective items in a sample? Answe

r(a) Mutually exclusive, so

P(E1 or E2) = P(E1) + P(E2) = 0.34 + 0.46 = 0.8

(b) P(more than 1) = 1 − 0.8 = 0.2

04/07/2023VSR 61

Q.18. Find the mean and variance of the random variable X, where probability distribution is given by the following table: X -2 -1 0 1 2 3

P(X) 0.10 0.20 0.30 0.20 0.15 0.05

AnswerMean = XP(X)

- 2 0.10 + -1 0.20 + 0 0.30 + 1 0.20 + 2 0.15 + 3 0.05. = 0.25.

Variance = X2P(X) – (mean)2

= - 22 0.10 + -12 0.20 + 02 0.30 + 12 0.20 + 22 0.15 + 32 0.05 – (0.25)2 = 1.85 – 0.0625 = 1.7875

04/07/2023VSR 62

Q.19. The SD of a binomial distribution (q + p)16 is 2 , its mean is ? Answe

rSD = npq = 2 ., squaring npq = 2 .

But n = 16 ; we know q = 1 – p

16p(1 – p) = 4 ; 4p(1 – p) = 2

Þ 4p2 – 4p + 1 = 0 Þ (2p – 1)2 = 0 ; p = ½

We know mean = np = 16 ½ = 8.

04/07/2023VSR 63

Q.20. In a binomial distribution mean = 5 and variance = 4 , then the number of trials is ?

Answer

Mean = np = 5. ; Variance = npq = 4 .

\npq/np = 4/5 q = 4/5.

\ p = 1 – q = 1 – 4/5 = 1/5

Now, n p = 5

i.e., n 1/5 = 5

n = 25

04/07/2023VSR 64

Q.21.How many words can be formed from the letter of the word “COMMITTEE”. Answe

r

We have the formulae !!!

!

321 nnn

n

Total no. of letters = 9 ; M = 2, T = 2, E = 2

3321 )!2(

!9

!2!.2!.2

!9

!!!

!

nnn

n

04/07/2023VSR 65

Q.22. In a leap year , the probability of having 53 Friday or Saturday is ? Answe

rIn a non-leap year there are 365 days, 52 complete weeks and 1 day.

That can be Monday, Tuesday, Wednesday, Thursday, Friday and Saturday.

P(Friday) = 1/7 ; P( Saturday) = 1/7

The required probability = 1/7 + 1/7 = 2/7 .

04/07/2023VSR 66

Q.23. A card is chosen at random from a deck of 52 cards. It is then replaced and a second card is chosen. What is the probability of choosing a Jack and an Eight.

Answer

P(Jack) = 4/52 ; P(8) = 4/52

Multiplication rule

P(Jack and Eight) = 4/52 4/52 = 16/2704 = 1/169.

04/07/2023VSR 67

Q.23. Obtain binomial distribution, if n = 6, p = 1/5

Answer

We have p = 1/5 and n = 6.

We know q = 1 – p = 1 – 1/5 = 4/5 .

Binomial distribution = (q + p)n

=

6

5

1

5

4

04/07/2023VSR 68

Q.24. The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. What is the probability that out of 5 such bulbs(i) None ; (ii) not more than one ; (iii) at least one will fuse after 150 days of use. Answe

rLet X represent the number of bulbs that will fuse after 150 days of use in an experiment of 5 trials. The trials are Bernoulli trials.

It is given that, p = 0.05 ; q = 1 – p = 1 – 0.05 = 0.95.

X has a binomial distribution with n = 5 and p = 0.05

04/07/2023VSR 69

(i) P (none) = P(X = 0)

Using this formulae

04/07/2023VSR 70

(ii) P (not more than one) = P(X ≤ 1)

04/07/2023VSR 71

(iii) P (at least one) = P(X ≥ 1)

04/07/2023VSR 72

The End of Probability

The End of Probability

The End of Probability

The End of Probability

The End of Probability

The End of Probability

The End of Probability

The End of Probability

The End of Probability

The End of Probability

The End o

f Pro

bability