proba

1

INTRODUCTION, SAMPLE SPACES & EVENTS Probability Let E be a random experiment (where we ‘know’ all possible outcomes but can’t predict what the particular outcome will be when the experiment is conducted). The set of all possible outcomes is called a sample space for the random experiment E. Example 1: Let E be the random experiment:

Toss two coins and observe the sequence of heads and tails. A sample space for this experiment could be { }TTHTTHHHS ,,,= . If however we only observe the number

of heads got, the sample space would be S = {0, 1, 2}.

Example 2: Let E be the random experiment:

Toss two fair dice and observe the two numbers on the top. A sample space would be

( ) ( ) ( ) ( )( ) ( ) ( )( )

( ) ��

�

��

�

�

��

�

��

�

�

−−−−−−−−−−

−−−−−−−−−−−

=

)6,6(,1,6|

,1,3

,3,2,2,2,1,26,1,,3,1,2,1,1,1

S

If however, we are interested only in the sum of the two numbers on the top, the sample space could be S = { 2, 3, …, 12}.


Count the number of machines produced by a factory until a defective machine is produced. A sample space for this experiment could be { }−−−−−−= ,3,2,1S .

2


Count the life length of a bulb produced by a factory. Here S will be { } ).,0[0| ∞=≥tt

Events

An event is a subset of the sample space.

Example 5: Suppose a balanced die is rolled and we observe the number on the top. Let A be the event: an even number occurs.

Thus in symbols,

{ } { }6,5,4,3,2,16,4,2 =⊂= SA

Two events are said to be mutually exclusive if they cannot occur together; that is there is no element common between them. In the above example if B is the event: an odd number occurs, i.e. B = { }5,3,1 , then A and

B are mutually exclusive.

Solved Examples Example 1: A manufacturer of small motors is concerned with three major types of defects. If A is the event that the shaft size is too large, B is the event that the windings are improper and C is the event that the electrical connections are unsatisfactory, express in words what events are represented by the following regions of the Venn diagram given below:

(a) region 2 (b) regions 1 and 3 together (c) regions 3, 5, 6 and 8 together.

3

Solution:

(a) Since this region is contained in A and B but not in C, it represents the event that the shaft is too large and the windings improper but the electrical connections are satisfactory.

(b) Since this region is common to B and C, it represents the event that the windings are improper and the electrical connections are unsatisfactory. (c) Since this is the entire region outside A, it represents the event that the shaft size is not too large.

Example 2: A carton of 12 rechargeable batteries contain one that is defective. In how many ways can the inspector choose three of the batteries and

(a) get the one that is defective (b) not get the one that is defective.

Solution:

(a) one defective can be chosen in one way and two good ones can be chosen in

55211

=��

��

ways. Hence one defective and two good can be chosen in 1 x 55 = 55

ways.

(b) Three good ones can be chosen in 165311

=��

��

ways

8

A 7

B 2 5 1

4 3 C 6

4

AXIOMS OF PROBABILITY Let E be a random experiment. Suppose to each event A, we associate a real number P(A) satisfying the following axioms: (i) ( ) 10 ≤≤ AP

(ii) ( ) 1=SP

(iii) If A and B are any two mutually exclusive events, then ( ) ( ) ( )BPAPBAP +=∪

(iv) If {A1, A2 - - - - - -An , …} is a sequence of pair- wise mutually exclusive events, then ...)(...)()(...)...( 2121 ++++=∪∪∪∪ nn APAPAPAAAP

We call P(A) the probability of the event A. Axiom 1 says that the probability of an event is always a number between 0 and 1. Axiom 2 says that the probability of the certain event S is 1. Axiom 3 says that the probability is an additive set function. Some elementary consequences of the Axioms 1. ( ) 0=φP

Proof: S= φ∪S .. Now S and φ are disjoint.

Hence .0)()()()( =�+= φφ PPSPSP Q.E.D.

2. If nAAA ,...,, 21 are any n pair-wise mutually exclusive events, then

( ) ( )�=

=∪∪∪n

iin APAAAP

121 ... .

Proof: By induction on n. Def.: If A is an event A′

� the complementary event = S-A (It is the shaded portion in the figure below)

A

5

3. )(1)( APAP −=′

Proof: AAS ′∪= Now )()()( APAPSP ′+= as A and A′ are disjoint or 1 = )()( APAP ′+ .

Thus )(1)( APAP −=′ . Q.E.D.

4. Probability is a

subtractive set function; i.e. If BA ⊂ , then

)()()( APBPABP −=− .

5. Probability is a monotone set function:

i.e. )()( BPAPBA ≤�⊂

Proof: ( )ABAB −∪= where A, B-A are disjoint.

Thus ).()()()( APABPAPBP ≥−+=

BA ∩ 6. If A, B are any two events,

( ) ( )BAPBPAPBAP ∩−+=∩ )()(

Proof: ( ) ( )BAABA ∩′∪=∪ )

where A and BA ∩′ are disjoint Hence ( ) ( )BAPAPBAP ∩′+=∪ )(

But ( ) ( ),BABAB ∩′∪∩=

union of two disjoint sets ( ) ( )

( ) ( ) ( ).)(

BAPBPBAPor

BAPBAPBP

∩−=∩′∩′+∩=�

( ) ( )BAPBPAPBAP ∩−+=∪∴ )()( . Q.E.D.

7. If A, B, C are any three events,

( ) )CBA(P)AC(P)CB(P)BA(P)C(P)B(P)A(PCBAP ∩∩+∩−∩−∩−++=∪∪ .

B

A

A B

BA ∩′

6

Proof:

( )( )

)CBA(P)CB(P)CA(P)BA(P)C(P)B(P)A(P))CB()CA((P)BA(P)C(P)B(P)A(P

)C)BA((P)C(P)BA(P)B(P)A(PCBAP)C(P)BA(P)CBA(P

∩∩+∩−∩−∩−++=∩∪∩−∩−++=

∩∪−+∩−+=∩∪−+∪=∪∪

More generally,

8. If nAAA ,...,, 21 are any n events.

)AAA(P)1(

...)AAA(P)AA(P)A(P

)A...AA(P

n211n

nkji1kji

nj1iji

n

1iI

n21

∩−−−−−−−∩∩−+

−∩∩+∩−=

∪∪∪

−

<≤<≤≤<≤=��

Finite Sample Space (in which all outcomes are equally likely) Let E be a random experiment having only a finite number of outcomes. Let all the (finite no. of) outcomes be equally likely. If { }naaaS ,...,, 21= ( naaa ,...,, 21 are equally likely outcomes), { } { } { }n21 a.......aaS ∪= .a

union of m.e. events. Hence { } { }( )naPaPaPSP −−−+= 21})({)(

But P({a1})=P({a2})= …= P({an}) = p (say) Hence 1 = p+ p+ . . . +p (n terms) or p = 1/n Hence if A is a subset consisting of ‘k’ of these outcomes,

A ={a1, a2………ak}, then nk

AP =)( = outcomesofno.Total

outcomesfavorableofNo..

7

Example 1: If a card is drawn from a well-shuffled pack of 52 cards find the probability of drawing

(a) a red king Ans: 522

(b) a 3, 4, 5 or 6 Ans: 5216

(c) a black card Ans: 21

(d) a red ace or a black queen Ans: 524

Example 2: When a pair of balanced die is thrown, find probability of getting a sum equal to

(a) 7.

Ans: 61

366 = (Total number of equally likely outcomes is

36 & the favourable number of outcomes = 6, namely (1,6), (2,5),, …(6,1).)

(b) 11 Ans: 362

(c) 7 or 11 Ans: 368

(d) 2, 3 or 12 Ans: = 364

361

362

361 =++ .

Example 3: 10 persons in a room are wearing badges marked 1 through 10. 3 persons are chosen at random and asked to leave the room simultaneously and their badge nos are noted. Find the probability that

(a) the smallest badge number is 5. (b) the largest badge number is 5.

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Solution:

(a) 3 persons can be chosen in 10C3 equally likely ways. If the smallest badge number is to be 5, the badge numbers should be 5 and any two of the 5 numbers 6, 7, 8, 9,10. Now 2 numbers out of 5 can be chosen in 5C2 ways. Hence the probability that the smallest badge number is 5 is 5C2 /10C3 .

(b) Ans. 4C2 /10C3 . Example 4: A lot consists of 10 good articles, 4 articles with minor defects and 2 with major defects. Two articles are chosen at random. Find the probability that

(a) both are good Ans: 2

162

10

CC

(b) both have major defects Ans: 2

162

2

CC

(c) At least one is good Ans: 1 – P(none is good) =2

1

166

1cc

−

(d) Exactly one is good Ans: 2

11

166.10

ccc

(e) At most one is good Ans. P(none is good) + P(exactly one is good) =

2

11

2

2

166.10

166

ccc

cc

+

(f) Neither has major defects Ans:2

2

1614

cc

(g) Neither is good Ans:2

2

166

cc

9

Example 5: From 6 positive and 8 negative integers, 4 integers are chosen at random and multiplied. Find the probability that their product is positive. Solution: The product is positive if all the 4 integers are positive or all of them are negative or two of them are positive and the other two are negative. Hence the probability is

��

��

��

��

��

��

+

��

��

��

��

+

��

��

��

��

414

28

26

41448

41446

Example 6: If, A, B are mutually exclusive events and if P(A) = 0.29, P(B) = 0.43, then

(a) 0.710.291)AP( =−=′

(b) P(A∪B) = 0.29 + 0.43 = 0.72 (c) P ( ) m.e.) are B and A since,B of subseta isA (as [0.29BA ′==′∩ )A(P

(d) 0.280.721B)P(A1)BAP( =−=∪−=′∩′

Example 7: P(A) = 0.35, P(B) = 0.73, P 0.14B)(A =∩ . Find

(a) P (A ∪ B) = P(A) + P(B) - P( A ∩ B) = 0.94. (b) 0.59B)P(AP(B)B)A(P =∩−=∩′

(c) 0.21B)P(AP(A))B(AP =∩−=′∩

(d) 0.860.141B)P(A1)B AP( =−=∩−=′∪′

Example 8: A, B, C are 3 mutually exclusive events. Is this assignment of probabilities possible?

P(A) = 0.3, P(B) = 0.4, P(C) = 0.5

10

Ans. P(A ∪ B ∪ C) = P(A) + P(B) + P(C) >1�NOT POSSIBLE

Example 9: Three newspapers are published in a city. A recent survey of readers indicated the following:

20% read A 8% read A and B 2% read all 16% read B 5% read A and C 14% read C 4% read B and C

Find probability that an adult chosen at random reads

(a) none of the papers.

Ans. 0.65100

24581416201C) BP(A 1 =��

��

� +−−−+++−=∪∪−

(b) reads exactly one paper.

P (Reading exactly one paper)

0.22100

769 =++=

(c) reads at least A and B given he reads at least one of the papers.

P (At least reading A and B given he reads at least one of the papers)

= 358

C)BP(AB)P(A =∪∪

∩

A B

C

9 6

3 6

2 2 7

11

CONDITIONAL PROBABILITY Let, A, B be two events. Suppose P(B) ≠ 0. The conditional probability of A occurring given that B has occurred is defined as

P(A | B) = probability of A given B = .P(B)

B)P(A ∩

Similarly we define P(B | A) = P(A)

B)P(A ∩ if P(A) ≠ 0.

Hence we get the multiplication theorem

0)P(A)(if )P(A).P(B/AB)P(A ≠=∩ )

0)P(B)(if )P(B).P(A/B ≠=

Example 10 A bag contains 4 red balls and 6 black balls. 2 balls are chosen at random one by one without replacement. Find the probability that both are red. Solution Let A be the event that the first ball drawn is red, B the event the second ball drawn is red. Hence the probability that both balls drawn are red =

152

93

104

A)|P(BP(A)B)P(A =×=×=∩

Independent events: Definition: We say two events A, B are independent if P(A ∩ B) = P(A). P(B) Equivalently A and B are independent if P(B | A) = P(B) or P(A | B) = P(A) Theorem If, A, B are independent, then

(a) A′ , B are independent (b) A, B′ are independent (c) B,A ′′ are independent

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Proof B)A( B)(A B ∩′∪∩=

Mutually exclusive

B)AP(B)P(AP(B) ∩′+∩=�

B)P(A-P(B)B)AP( ∩=∩′�

= P(B) – P(A) (P/B) = P(B) [1-P(A)] = P(B) P( )A′

∴A, B′ are also independent. By the same reasoning, A′ and B are independent. So again A′ and B′ are independent.

Example 11 Find the probability of getting 8 heads in a row in 8 tosses of a fair coin. Solution If Ai is the event of getting a head in the ith toss, A1, A2, …, A8 are independent and

P(Ai) = 21

for all i. Hence P(getting all heads) =

P(A1) P(A2)…P(An) = 8

21�

��

Example 12 It is found that in manufacturing a certain article, defects of one type occur with probability 0.1 and defects of other type occur with probability 0.05. Assume independence between the two types of defects. Find the probability that an article chosen at random has exactly one type of defect given that it is defective.

A B A∩B

BA ∩′

13

Let A be the event that article has exactly one type of defect.

Let B be the event that the article is defective.

Required P(B)

B)P(AB)|P(A

∩=

P(B) = P(D ∪ E) where D is the event it has type one defect

E is the event it has type two defect

= P(D) + P(E) – P(D ∩ E) = 0.1 + 0.05 - (0.1) (0.05) = 0.145

P(A ∩ B) = P (article is having exactly one type of defect) = P(D) + P(E) – 2 P(D ∩ E) = 0.1 + 0.05 - 2 (0.1) (0.05) = 0.14

∴Probability = 145.014.0

[Note: If A and B are two events, probability that exactly only one of them occurs is P(A) + P(B) – 2P(A ∩ B)]

Example 13 An electronic system has 2 subsystems A and B. It is known that

P (A fails) = 0.2

P (B fails alone) = 0.15

P (A and B fail) = 0.15

Find (a) P (A fails | B has failed) (b) P (A fails alone)

14

Solution

(a) P(A fails | B has failed) 21

0.300.15

failed)P(Bfailed)B andP(A ===

(b) P (A fails alone) = P (A fails) – P (A and B fail) = 0.02-0.15 = 0.05

Example 14 A binary number is a number having digits 0 and 1. Suppose a binary number is made up of ‘n’ digits. Suppose the probability of forming an incorrect binary digit is p. Assume independence between errors. What is the probability of forming an incorrect binary number?

Ans 1- P (forming a correct no.) = 1 – (1-p)n . Example 15 A question paper consists of 5 Multiple choice questions each of which has 4 choices (of which only one is correct). If a student answers all the five questions randomly, find the probability that he answers all questions correctly.

Ans 5

41�

��

.

Theorem on Total Probability

Let B1, B2, …, Bn be n mutually exclusive events of which one must occur. If A is any other event, then

( ) )BP(A......BAP)BP(AP(A) 21 n∩++∩+∩=

)B|P(A)P(B� ii1i==

n

(For a proof, see your text book.) Example 16

There are 2 urns. The first one has 4 red balls and 6 black balls. The second has 5 red balls and 4 black balls. A ball is chosen at random from the 1st and put in the 2nd. Now a ball is drawn at random from the 2nd urn. Find the probability it is red.

15

Solution:

Let B1 be the event that the first ball drawn is red and B2 be the event that the first ball drawn is black. Let A be the event that the second ball drawn is red. By the theorem on total probability,

P(A) = P(B1) P(A | B1) + P(B2) P(A | B2) = 10054

105

106

106

104 =×+× =0.54.

Example 17: A consulting firm rents cars from three agencies D, E, F. 20% of the cars are rented from D, 20% from E and the remaining 60% from F. If 10% of cars rented from D, 12% of cars rented from E, 4% of cars rented from F have bad tires, find the probability that a car rented from the consulting firm will have bad tires.

Ans. (0.2) (0.1) + (0.2) (0.12) + (0.6) (0.04) Example 18: A bolt factory has three divisions B1, B2, B3 that manufacture bolts. 25% of output is from B1, 35% from B2 and 40% from B3. 5% of the bolts manufactured by B1 are defective, 4% of the bolts manufactured by B2 are defective and 2% of the bolts manufactured by B3 are defective. Find the probability that a bolt chosen at random from the factory is defective.

Ans. 100

210040

1004

10035

1005

10025 ×+×+×

16

BAYES’ THEOREM Let B1, B2, ……….Bn be n mutually exclusive events of which one of them must occur. If A is any event, then

)B|)P(A P(B�

)B|)P(A P(BP(A)

)BP(AA)| P(B

ii1i

kkkk n

=

=∩

=

Example 19 Miss ‘X’ is fond of seeing films. The probability that she sees a film on the day before the test is 0.7. Miss X is any way good at studies. The probability that she maxes the test is 0.3 if she sees the film on the day before the test and the corresponding probability is 0.8 if she does not see the film. If Miss ‘X’ maxed the test, find the probability that she saw the film on the day before the test.

Solution Let B1 be the event that Miss X saw the film before the test and let B2 be the complementary event. Let A be the event that she maxed the test.

Required. P(B1 | A)

)B|P(A P(B))B|P(A )P(B)B|)P(A P(B

211

11

×+×=

Example 20 At an electronics firm, it is known from past experience that the probability a new worker who attended the company’s training program meets the production quota is 0.86. The corresponding probability for a new worker who did not attend the training program is 0.35. It is also known that 80% of all new workers attend the company’s training

8.03.03.07.03.07.0

×+××=

17

program. Find probability that a new worker who met the production quota would have attended the company’s training programme. Solution Let B1 be the event that a new worker attended the company’s training programme. Let B2 be the complementary event, namely a new worker did not attend the training programme. Let A be the event that a new worker met the production quota. Then we

want P(B1 | A) = 35.02.086.08.0

8.08.0×+×

×.

Example 21 A printing machine can print any one of n letters L1, L2,……….Ln. It is operated by electrical impulses, each letter being produced by a different impulse. Assume that there is a constant probability p that any impulse prints the letter it is meant to print. Also assume independence. One of the impulses is chosen at random and fed into the machine twice. Both times, the letter L1 was printed. Find the probability that the impulse chosen was meant to print the letter L1. Solution: Let B1 be the event that the impulse chosen was meant to print the letter L1. Let B2 be the complementary event. Let A be the event that both the times the letter L1 was printed.

P(B1) = n1

. P(A|B1) = p2. Now the probability that an impulse prints a wrong letter is (1-

p). Since there are n-1 ways of printing a wrong letter, P(A|B2) = 1

1−

−n

p. Hence P(B1|A)

= )B|P(A)P(B)B|P(A)P(B

)B|P(A)P(B

2211

11

×+××

22

2

111

11

1

�

��

−−

�

��

−+

�

��

=

np

np

n

pn . This is the required probability.

18

Miscellaneous problems 1 (a). Suppose the digits 1,2,3 are written in a random order. Find probability that at

least one digit occupies its proper place.

Solution There are 3! = 6 ways of arranging 3 digits (See the figure), out of which in 4 arrangements , at least one digit occupies its proper place. Hence the probability is

3!4

=64

. 123 213 312

132 231 321 (Remark. An arrangement like 231, where no digit occupies its proper place is called a derangement.)

(b) Same as (a) but with 4 digits 1,2,3,4 Ans. 2415

(Try proving this.)

Solution Let A1 be the Event 1st digit occupies its proper place

A2 be the Event 2nd digit occupies its proper place A3 be the Event 3rd digit occupies its proper place

A4 be the Event 4th digit occupies its proper place

P(at least one digit occupies its proper place) =P(A1∪A2 ∪A3 ∪A4) =P(A1) + P(A2) + P(A3) + P(A4)

(There are 4C1 terms each with the same probability) )A A P(...)AP(A)AP(A)AP(A 43413121 ∩−−∩−∩−∩−

(There are 4C2 terms each with the same probability) )AAP(A ...)AAP(A ).AAP(A 4324 21321 ∩∩++∩∩+∩∩+

(There are 4C3 terms each with the same probability) - )AAAA P( 4321 ∩∩∩

4!0!

4c4!1!

4c4!2!

4c4!3!

4c 4321 −+−=

19

241

61

21

1 −+−=

2415

24141224 =−+−=

(c) Same as (a) but with n digits.

Solution

Let A1 be the Event 1st digit occupies its proper place A2 be the Event 2nd digit occupies its proper place

…………………… An be the Event nth digit occupies its proper place

P(at least one digit occupies its proper place)

= P(A1∪A2 ∪ … ∪An)

= !

1(-1)...... -

n!3)!(n

ncn!

2)!(nnc

n!1)!(n

nc 1-n321 n

+−+−−−

!

11)(..........

4!1

3!1

2!1

1 1n

n−−−+−= ≈ !e1 −− (for n large).

2. In a party there are ‘n’ married couples. If each male chooses at random a

female for dancing, find the probability that no man chooses his wife.

Ans 1-(!

11)(..........

4!1

3!1

2!1

1 1n

n−−−+− ).

3. A and B play the following game. They throw alternatively a pair of dice.

Whosoever gets sum of the two numbers on the top as seven wins the game and the game stops. Suppose A starts the game. Find the probability (a) A wins the game (b) B wins the game.

20

Solution A wins the game if he gets seven in the 1st throw or in the 3rd throw or in the

5th throw or …. Hence P(A wins) = 61

65

65

65

65

61

65

65

61 ××××+××+ + …

= .116

362536

61

65

1

61

2 =−

=

�

��

− P(B wins) = complementary probability =

115

.

4. Birthday Problem

There are n persons in a room. Assume that nobody is born on 29th Feb. Assume that any one birthday is as likely as any other birth day. Find the probability that no two persons will have same birthday.

Solution

If n > 365, at least two will have the same birthday and hence the probability that no two will have the same birthday is 0.

If n ≤ 365, the desired probability is ( )[ ]

n(365)1n365.........364365 −−××= .

5. A die is rolled until all the faces have appeared on top.

(a) What is probability that exactly 6 throws are needed?

Ans. 66!6

(b) What is probability that exactly ‘n’ throws are needed? ( )6n >

21

6. Polya’s urn problem

An urn contains g green balls and r red balls. A ball is chosen at random and its color is noted. Then the ball is returned to the urn and c more balls of same color are added. Now a ball is drawn. Its color is noted and the ball is replaced. This process is repeated.

(a) Find probability that 1st ball drawn is green.

Ans. rg

g+

(b) Find the probability that the 2nd ball drawn is green.

Ans. rg

gcrg

grg

rcrg

cgrg

g+

=+++

+++

+×+

(c) Find the probability that the nth ball drawn is green.

The surprising answer is rg

g+

.

7. There are n urns and each urn contains a white and b red balls. A ball is chosen from Urn 1 and put into Urn 2. Now a ball is chosen at random from urn 2 and put into urn 3 and this is continued. Finally a ball drawn from Urn n. Find the probability that it is white. Solution

Let pr = Probability that the ball drawn from Urn r is white.

∴1

)p(11

1pp 1r1rr ++

×−+++

+×= −− aaa

baa

; r = 1, 2, …, n.

This is a recurrence relation for pr. Noting that p1 = ba

a+

, we can find pn.

22

MATHEMATICAL EXPECTATION & DECISION MAKING

Suppose we roll a die n times. What is the average of the n numbers that appear on the top?

Suppose 1 occurs on the top n1 times Suppose 2 occurs on the top n2 times Suppose 3 occurs on the top n3 times Suppose 4 occurs on the top n4 times Suppose 5 occurs on the top n5 times Suppose 6 occurs on the top n6 times

Total of the n numbers on the top = 621 n....6..........n1n1 ×+×+×

∴Average of the n numbers,

nnnn621621 n

6...n

2n

1n6..........n2n1

×++×+×=××+×

=

Here clearly n1, n2, …, n6 are unknown. But by the relative frequency definition of

probability, we may approximate nn1 by P(getting 1 on the top) =

61

, nn2 by

P(getting 2 on the top) = 61

, and so on. So we can ‘expect’ the average of the n

numbers to be 5.327 = . We call this the Mathematical Expectation of the number

on the top.

Definition

Let E be a random experiment with n outcomes a1, a2 ……….an. Suppose P({a1})=p1, P({a2})=p2, …, P({an})=pn. Then we define the mathematical expectation as

nn2211 pa.........papa ×+×+×

23

Problems

1. If a service club sells 4000 raffle tickets for a cash prize of $800, what is the

mathematical expectation of a person who buys one of these tickets?

Solution. 2.051

)(04000

1800 ==×+×

2. A charitable organization raises funds by selling 2000 raffle tickets for a 1st prize

worth $5000 and a second prize $100. What is mathematical expectation of a person who buys one of the tickets?

Solution. )(02000

1100

20001

5000 ×+×+×

3. A game between 2 players is called fair if each player has the same mathematical

expectation. If some one gives us $5 whenever we roll a 1 or a 2 with a balanced die, what we must pay him when we roll a 3, 4, 5 or 6 to make the game fair? Solution. If we pay $x when we roll a 3, 4, 5, or 6 for the game to be fair,

62

564 ×=×x or x = 10. That is we must pay $10.

4. Gambler’s Ruin

A and B are betting on repeated flips of a balanced coin. At the beginning, A has m dollars and B has n dollars. After each flip the loser pays the winner 1 dollar and the game stops when one of them is ruined. Find probability that A will win B’s n dollars before he loses his m dollars. Solution. Let p be the probability that A wins (so that 1-p is the probability that B wins). Since the game is fair, A’s math exp = B’s math exp.

Thus ( ) 0.pp)m(1p1 0pn +−=−+× or nm

mp

+=

24

5. An importer is offered a shipment of machines for $140,000. The probability that he will sell them for $180,000, $170,000 (or) $150,000 are respectively 0.32, 0.55, and 0.13. What is his expected profit?

Solution. Expected profit

= 13.0000,1055.0000,3032.0000,40 ×+×+×

=$30,600 6. The manufacturer of a new battery additive has to decide whether to sell her

product for $80 a can and for $1.2 a can with a ‘double your money back if not satisfied’ guarantee. How does she feel about the chances that a person will ask for double his/her money back if

(a) she decides to sell the product for $0.80 (b) she decides to sell the product for $1.20 (c) she can not make up her mind?

Solution. In the 1st case, she gets a fixed amount of $0.80 a can

In the 2nd case, she expects to get for each can (1.20) (1-p) + (-1.2) (p) = 1.20 – (2.4) p

Let p be the prob that a person will ask for double his money back. (a) happens if 0.80 > 1.20 –2.40 p

� p > 1/6 (b) happens if

p < 1/6 (c) happens if p = 1/6

25

7. A manufacturer buys an item for $1.20 and sells it for $4.50. The probabilities for a demand of 0, 1, 2, 3, 4, “5 or more” items are 0.05, 0.15, 0.30, 0.25, 0.15, 0.10 respectively. How many items he must stock to maximize his expected profit?

No. of items stocked No. sold with prob. Exp. profit 0 0 1 0

1 0 0.05 1 0.95

175.21.295.0

5.405.00

=−×

+×

2 0 0.05 1 0.15 2 0.80 675.3

2.480.0915.05.405.00

=−×+

×+×

3

0 0.05 1 0.15 2 0.30 3 0.50 ��=

−×+×+

×+×

3.615.05.1330.09

15.05.405.00

4 2.85

5 0.525

6 0.45

Hence he must stock 3 items to maximize his expected profit.

8. A contractor has to choose between 2 jobs. The 1st job promises a profit of $240,000 with probability 0.75 and a loss of $60,000 with probability 0.25. The 2nd job promises a profit of $360,000 with probability 0.5 and a loss of $90,000 with probability 0.5.

(a) Which job should the contractor choose to maximize his expected profit?

i. Exp. profit for job1 = 000,15541

000,6043

000,240 =×−×

ii. Exp. profit for job2 = 36,000 000,13521

000,9021 =×−×

Go in for job1.

(b) What job would the contractor probably choose if her business is in bad shape and she goes broke unless, she makes a profit of $300,000 on her next job.

Ans:- She takes the job2 as it gives her higher profit.

26

RANDOM VARIABLES Let E be a random experiment. A random variable (r.v) X is a function that associates to each outcome s, a unique real number X (s). Example 1 Let E be the random experiment of tossing a fair coin 3 times. We see that there are

823 = outcomes TTT, HTT, THT, TTH, HHT, HTH, THH, HHH all of which are equally likely. Let X be the random variable that ‘counts’ the number of heads obtained. Thus X can take only 4 values 0,1,2,3. We note that

( ) ( ) ( ) ( ) .81

3,83

2,83

1,81

0 ======== XPXPXPXP This is called the

probability distribution of the rv X. Thus the probability distribution of a rv X is the listing of the probabilities with which X takes all its values. Example 2 Let E be the random experiment of rolling a pair of balanced die. There are 36 possible equally likely outcomes, namely (1,1), (1,2)…… (6,6). Let X be the rv that gives the sum of the two nos on the top. Hence X take 11 values namely 2,3……12. We note that the probability distribution of X is

( ) ( ) ( ) ( )362

11XP3XP,361

12XP2XP ======== ,

( ) ( ) ,363

10XP4XP ====

( ) ( )364

9XP5XP ==== .

( ) ( ) ( ) .61

366

7XP,365

8XP6XP =======

Example 3 Let E be the random experiment of rolling a die till a 6 appears on the top. Let X be the no of rolls needed to get the “first” six. Thus X can take values 1,2,3…… Here X takes an infinite number of values. So it is not possible to list all the probabilities with which X takes its values. But we can give a formula.

27

( ) ( ).....2,161

65

1

=�

��

==−

xxXPx

(Justification: X = x means the first (x-1) rolls gave a number (other than 6) and

the xth roll gave the first 6. Hence ( ) )61

65

61

65

...65

65

1

1

−

−

�

��

=×××==x

timesx

xXP��

Discrete Random Variables We say X is a discrete rv of it can take only a finite number of values (as in example 1,2 above) or a “countably” infinite values (as in example 3). On the other hand, the annual rainfall in a city, the lifelength of an electronic device, the diameter of washers produced by a factory are all continuous random variables in the sense they can take (theoretically at least) all values in an ‘interval’ of the x-axis. We shall discuss continuous rvs a little later. Probability distribution of a Discrete RV Let X be a discrete rv with values ......, 21 xx

Let ( ) ( )( ).....2,1ixXPxf ii ===

We say that ( ){ } ....2,1iixf = is the probability distribution of the rv X.

Properties of the probability distribution

(i) ( ) .....2,1iallfor0xf i =≥

(ii) ( ) 1xf ii

=�

The first condition follows from the fact that the probability is always .0≥ The second condition follows from the fact that the probability of the certain event = 1.

28

Example 4 Determine whether the following can be the probability distribution of a rv which can take only 4 values 1,2,3 and 4.

(a) ( ) ( ) ( ) ( ) 26.0426.0326.0226.01 ==== ffff .

No as the sum of all the “probabilities” > 1. (b) ( ) ( ) ( ) ( ) 28.0429.03,28.0215.01 ==== ffff .

Yes as these are all 0≥ and add up to 1.

(c) ( ) 4,3,2,116

1 =+= xx

xf .

No as the sum of all the probabilities < 1. Binomial Distribution Let E be a random experiment having only 2 outcomes, say ‘success’ and ‘failure’. Suppose that P(success) = p and so P(failure) = q (=1-p). Consider n independent repetitions of E (This means the outcome in any one repetition is not dependent upon the outcome in any other repetition). We also make the important assumption that P(success) = p remains the same for all such independent repetitions of E. Let X be the rv that ’counts’ the number of successes obtained in n such independent repetitions of E. Clearly X is a discrete rv that can take n+1 values namely 0,1,2,….n. We note that there are

n2 outcomes each of which is a ‘string’ of n letters each of which is an S or F (if n =3, it will be FFF, SFF, FSF, FFS, SSF, SFS, FSS, SSS). X = x means in any such outcome there are x successes and (n-x) failures in some order. One such will be

xnxFFFFSSSS

−.... . Since all

the repetitions are independent prob of this outcome will be xnx qp − . Exactly the same

prob would be associated with any other outcome for which X = x. But x successes can

occur out of n repetitions in ��

��

xn

mutually exclusive ways. Hence

( ) ( )....n1,0,xqpxn

xXP xnx =��

��

== −

29

We say X has a Binomial distribution with parameters n ( ≡ the number of repetitions)

and p (Prob of success in any one repetition). We denote ( ) ( )p,n;xbxXP by= to show its dependence on x, n and p. The letter ‘b’

stands for binomial. Since all the above (n+1) probabilities are the (n+1) terms in the expansion of the

binomial ( )npq + , X is said to have a binomial distribution. We at once see that the sum

of all the binomial probabilities = ( ) .11 ==+ nnpq

The independent repetitions are usually referred to as the “Bernoulli” trials. We note that

( ) ( )q,n;xnbp,n;xb −=

(LHS = Prob of getting x successes in n Bernoulli trials = prob of getting n-x failures in n Bernoulli trials = R.H.S.) Cumulative Binomial Probabilities Let X have a binomial distribution with parameters n and p.

( ) ( ) P0XPxXP +==≤ ( ) ( )xXP......1X =+=

= ( )p,n;kbx

0k�

=

is denoted by ( )pnxB ,; and is called the cumulative Binomial distribution function. This

is tabulated in Table 1 of your text book. We note that

( ) ( ) ( ) ( )( ) ( )p,n;1xBp,n;xB

1xXPxXPxXpp,n;xb−−=

−≤−≤===

Thus ( )60.00,12;9b = ( ) ( )60.0,12;860.0,12;9 BB −

1419.0

7747.09166.0=

−=

(You can verify this by directly calculating ( )).9;12,0.60b

30

Example 5 (Exercise 4.15 of your book) During one stage in the manufacture of integrated circuit chips, a coating must be applied. If 70% of the chips receive a thick enough coating find the probability that among 15 chips.

(a) At least 12 will have thick enough coatings. (b) At most 6 will have thick enough coatings. (c) Exactly 10 will have thick enough coatings.

Solution Among 15 chips, let X be the number of chips that will have thick enough coatings. Hence X is a rv having Binomial distribution with parameters n =15 and p = 0.70.

(a) ( ) ( )11XP112XP ≤−=≥

( )

3969.07031.0170.0,15;111

=−=−= B

(b) ( ) ( )70.0,15;6B6XP =≤

0152.0= (c) ( ) ( ) ( )70.0,15;9B70.0,15;10B10XP −==

2065.0

2784.04849.0=

−=

Example 6 (Exercise 4.19 of your text book) A food processor claims that at most 10% of her jars of instant coffee contain less coffee than printed on the label. To test this claim, 16 jars are randomly selected and contents weighed. Her claim is accepted if fewer than 3 of the 16 jars contain less coffee (note that 10% of 16 = 1.6 and rounds to 2). Find the probability that the food processor’s claim will be accepted if the actual percent of the jars containing less coffee is (a) 5% (b) 10% (c) 15% (d) 20% Solution: Let X be the number of jars that contain less coffee (than printed on the label) (among the 16 jars randomly chosen. Thus X is a random variable having a Binomial distribution

31

with parameters n = 16 and p (the prob of “success” = The prob that a jar chosen at random will have less coffee)

(a) Here p = 5% = 0.05 Hence P (claim is accepted) = ( ) ( ) .9571.005.0,16;2B2XP ==≤

(b) Here p = 10% = 0.10

Hence P (claim is accepted) = ( ) 7892.001.0,16;2 =B

(c) Here p = 15% = 0.15.

Hence P (claim is accepted) = B ( ) 5614.015.0,16;2 =

(d) Here p = 20% = 0.20

Hence P(claims accepted) = ( ) 3518.029.0,16,2 =B

Binomial Distribution – Sampling with replacement Suppose there is an urn containing 10 marbles of which 4 are white and the rest are black. Suppose 5 marbles are chosen with replacement. Let X be the rv that counts the no of white marbles drawn. Thus X = 0,1,2,3,4 or 5 (Remember that we replace each marble in the urn before drawing the next one. Hence we can draw 5 white marbles)

P (“Success”) = P (Drawing a white marble in any one of the 5 draws) = 104

(remember

we draw with replacement).

Thus X has a Binomial distribution with parameters n = 5 and 104=p

Hence ( ) �

��

==104

,5;xbxXP

Mode of a Binomial distribution We say 0x is the mode of the Binomial distribution with parameters n and p if

( )0xXP = is the greatest. From the binomial tables given in the book we can easily see

that

32

When ( ) .mod55,21

,10 etheisorgreatesttheisXPpn ===

Fact

( )( ) ( )pnpxif

pp

nxn

pnxbpnxb −−<>

−×

+−=+

1111;;

,;1

( )( )ppnnif

pnpxif

−−><−−==

1111

Thus so long as x <np – (1-p) the binomial probabilities increase and if x> np-(1-p) they decrease. Hence if np-(1-p) = x0 is an integer, then the mode is .100 +xandx If n – (1-p)

in not an integer and if 0x = smallest integer ( )pnp −−≥ 1 , the mode is .x 0

Hypergeometric Distribution (Sampling without replacement) An urn contains 10 marbles of which 4 are white. 5 marbles are chosen at random without replacement. Let X be the rv that counts the number of white marbles drawn. Thus X can take 5 values names 0,1,2,3,4. What is P (X = x)? Now out of 10 marbles 5

can be chosen in ��

��

510

equally like ways, out of which there will be ��

��

−��

��

xx 564

ways of

drawing x white marbles (and so 5-x read marbles) (Reason out of 4 white marbles, x can

be chosen in ��

��

x

4 ways and out of 6 red marbles, 5-x can be chosen in ��

��

− x56

ways).

Hence ( ) .4,3,2,1,0

5105

64

=

��

��

��

��

−��

��

== xxx

xXP

We generalize the above result. A box contains N marbles out of which a are white. n marbles are chosen without replacement. Let X be the random variable that counts the number of white marbles drawn. X can take the values 0,1,2……. n.

33

( ) ....2,1,0=

��

��

��

��

−−

��

��

== x

n

Nxn

aN

x

a

aXP n

(Note x must be less than or equal to a and n-x must be less than or equal to N-a) We say the rv X has a hypergeometric distribution with parameters n,a and N. We denote P(X=x) by h (x;n,a,N). Example 7 (Exercise 4.22 of your text book) Among the 12 solar collectors on display, 9 are flat plate collectors and the other three are concentrating collectors. If a person choses at random 4 collectors, find the prob that 3 are flat plate ones.

Ans ( )��

��

��

��

��

��

=

412

13

39

12,9,4;3h

Example 8 (Exercise 4.24 of your text book) If 6 of 18 new buildings in a city violate the building code, what is the probability that a building inspector, who randomly selects 4 of the new buildings for inspection, will catch

(a) None of the new buildings that violate the building code

Ans ( )��

��

��

��

=

4184

12

18,6,4;1h

(b) One of the new buildings that violate the building code

34

Ans ( )��

��

��

��

��

��

=

418

312

16

18,6,4;1h

(c) Two of the new buildings that violate the building code

Ans ( )��

��

��

��

��

��

=

418

212

26

18,6,4;2h

(d) At least three of the new buildings that violate the building code Ans ( ) ( )18,6,4;418,6,4;3 hh +

(Note: We choose 4 buildings out of 18 without replacement. Hence hypergeometric distribution is appropriate)

Binomial distribution as an approximation to the Hypergeometric Distribution We can show that ( ) ( ) ∞→→ NaspnxbNanxh ,;,,;

(Where )"" successaofprobNa

p == . Hence if N is large the hypergeometric

probability ( )N,a,n;xh can be approximated by the binomial probability

( )p,n;xb where .Na

p =

Example 9 (exercise 4.26 of your text)

A shipment of 120 burglar alarms contains 5 that are defective. If 3 of these alarms are randomly selected and shipped to a customer, find the probability that the customer will get one defective alarm.

(a) By using the hypergemetric distribution (b) By approximating the hypergeometric probability by a binomial probability.

35

Solution

Here N = 120 (Large!) a = 5 n = 3 x =1

(a) Reqd prob = ( )120,5,3;1h

1167.0280840

65555

3120

2115

15

=×=

��

��

��

��

��

��

=

(b) ( ) �

��

≈120

5,3;1120,5,3;1 bh

1148.0120

51

1205

13

2

=�

��

−�

��

�

��

=

Example 10 (Exercise 4.27 of your text) Among the 300 employees of a company, 240 are union members, while the others are not. If 8 of the employees are chosen by lot to serve on the committee which administrates the provident fund, find the prob that 5 of them will be union members while the others are not.

(a) Using hypergemoretric distribution (b) Using binomial approximation

Solution

Here N = 300, a = 240, n = 8 x = 5

(a) ( )300,240,8;5h

(b) �

��

≈300240

8,5;b

36

THE MEAN AND VARIANCE OF PROBABILITY DISTRIBUTIONS We know that the equation of a line can be written as .cmxy += Here m is the slope and

c is the y intercept. Different m,c give different lines. Thus m and c characterize a line. Similarly we define certain numbers that characterize a probability distribution. The mean of a probability distribution is simply the mathematical expectation of the corresponding r.v. If a rv X takes on the values .....xx 2,1 with probabilities

( ) ( )....,xf,xf 21 its mathematical expectation or expected value is

( ) ( ) ( ) obabilityvaluexxPxxxfxxfx iii

Pr......2211 ×===++ ��

We use the symbol µ to denote the mean of X.

Thus ( ) ( )ii xxPxXE === �µ (Summation over all xi in the Range of X)

Example 11 Suppose X is a rv having the probability distribution

X 1 2 3

Prob 21

31

61

Hence the mean µ of the prob distribution (of X) is

35

61

331

221

1 =×+×+×=µ

Example 12 Let X be the rv having the distribution

X 0 1

Prob q p

37

where .10Thus.1 ppqpq =×+×=−= µ

The mean of a Binomial Distribution Suppose X is a rv having Binomial distribution with parameters n and p. Then Mean of .npX =µ=

(Read the proof on pages 107-108 of your text book) The mean of a hypergeometric Distribution

If X is a rv having hypergeometric distribution with parameters .,,,Na

nthenanN =µ

Digression The mean of a rv x give the “average” of the values taken by the rv. X. Thus the average marks in a test is 40 means the students would have got marks less than 40 and greater than 40 but it averages out to be 40. But we do not get an idea about the spread ( ≡ deviation from the mean) of the marks. This spread is measured by the variance. Informally speaking by the average of the squares of deviation from the mean. Variance of a Probability Distribution of X is defined as the expected value of

( )2X µ−

Variance of 2X σ=

( ) ( )Xi

i2

i

Rx

xXP�x

∈

=−=�

Note that R.H.S is always 0≥ (as it is the sum of non-ve numbers)

The positive square root 2of σσ is called the standard deviation of X and has the

same units as X and .µ

38

Example 13 For the rv X having the prob distribution given in example 11, the variance is

95

61

916

31

91

21

94

61

35

331

35

221

35

1222

=×+×+=

×�

��

−+×�

��

−+×�

��

−

x

We could have also used the equivalent formula

( ) ( )( )

.95

925

310

310

1860

69

34

21

61

331

221

1XEHere

XEXE

2

2222

2222

=−=σ∴

==++=×+×+×=

µ−=µ−=σ

Example 14 For the probability distribution of example 12,

( )( ) pqp1ppp

pp1qoXE22

222

=−=−=σ∴=×+×=

Variance of the Binomial Distribution

npq=2σ

Variance of the Hypergeometric Distribution

.1

.12

−−

�

��

−=N

nNNa

Na

nσ

39

CHEBYCHEV’S THEOREM

Suppose X is a rv with mean µ and variance 2σ . Chebychev’s theorem states that: If k

is a constant > 0,

( ) 2k1

k|X|P ≤σ≥µ−

In words the prob of getting a value which deviates from its mean µ by at least σk is at

most 2

1k

.

Note: Chebyshev’s Theorem gives us an upper bound of the prob of an event. Mostly it is of theoretical interest.

Example 15 (Exercise 4.44 of your text)

In one out of 6 cases, material for bullet proof vests fails to meet puncture standards. If 405 specimens are tested, what does Chebyshev theorem tell us about the prob of getting at most 30 or at least 105 cases that do not meet puncture standards?

Here 2

13561

405 =×== npµ

215

65

61

4052

=∴

××==

σ

σ qpn

Let X = no of cases out of 405 that do not meet puncture standards Reqd ( )105Xor30XP ≥≤

Now 2

75X30X −≤µ−�≤

275

X105X ≥µ−�≥

Thus σ=≥µ−�≥≤ 52

75|X|105Xor30X

40

( ) ( ) 04.0251

51

5|X|P105Xor30XP 2 ==≤σ≥µ−=≥≤∴

Example 16 (Exercise 446 of your text) How many times do we have to flip a balanced coin to be able to assert with a prob of at most 0.01 that the difference between the proportion of tails and 0.50 will be at least 0.04? Solution: Suppose we flip the coin n times and suppose X is the no of tails obtained. Thus the

proportion of tails = ��

��

=

flipsofNoTotaltailsofNo

nX

. We must find n so that

10.00.040.50nX

P ≤��

��

≥−

Now X = no of tails among n flips of a balanced coin is a rv having Binomial distribution with parameters n and 0.5. Hence ( ) 50.0nnpXE ×===µ

( )50.050.0 ==×== qpasnqpnσ

Now 04.050.0nX ≥− is equivalent to .n04.050.0nX ≥×−

We know ( ) 2k1

kXP ≤σ≥µ−

Here nk 04.0=σ

nn

nk 08.0

50.0

04.0 =×

=∴

41

( )

( )

( )15625

08.100

nor

.100n08.ifor10001.01

kif

01.0k1

k|X|P

04.050.0nX

P

2

22

2

=≥

≥=≥

≤≤σ≥µ−=

��

��

≥−∴

Law of large Numbers Suppose a factory manufactures items. Suppose there is a constant prob p that an item is defective. Suppose we choose n items at random and let X be the no of defectives found. Then X is a rv having binomial distribution with parameters n and p.

( ) npqiancevar,npXEmean 2 =σ==µ∴

Let ε be any no > 0.

Now ��

��

ε≥− pnX

P

( ) ( )σ≥µ−=ε≥−= kxPnnpXP (where ε=σ nk )

≤ .nas0npq

nnpq

n)theorems'Chebyshevby(

k1

22222

2

2 ∞→→ε

=ε

=ε

σ=

Thus we can say that the prob that the proportion of defective items differs from the actual prob. p by any + ve no ∞→→ nas0ε . (This is called the Law of Large

numbers) This means “most of the times” the proportion of defectives will be close to the actual

(unknown) prob p that an item is defective for large n. So we can estimate nX

byp , the

(Sample) proportion of defectives.

42

POISSON DISTRIBUTION A random variable X is said to have a Poisson distribution with parameter 0>λ if its probability distribution is given by

( ) ( ) ......2,1,0!

; ==== − xx

exfxXPxλλ λ

We can easily show: mean of λ=µ=X and variance of .X 2 λ=σ=

Also ( )xXP = is largest when λλ−λ= ifand1x is an integer and when [ ]λ=x = the

greatest integer λ≤ (when λ is not an integer). Also note that ( ) .0 ∞→→= xasxXP

POISSON APPROXIMATION TO BINOMIAL DISTRIBUTION Suppose X is a rv having Binomial distribution with parameters n and p. We can easily show ( ) ( ) ( ) ∞→→== nas�x;fxXPpn,x;b in such a way that np remains a constant

.λ Hence for n large, p small, the binomial prob ( )pnxb ,; can be approximated by the

Poisson prob ( )λ;xf where .np=λ

Example 17

( )03.0,100;3b

( )!33

3;333−

=≈ ef

Example 18 (Exercise 4.54 of your text) If 0.8% of the fuses delivered to an arsenal are defective, use the Poisson approximation to determine the probability that 4 fuses will be defective in a random sample of 400. Solution If X is the number of defectives in a sample of 400, X has the binomial distribution with parameters n = 400 and p = 0.8% = 0.008.

43

Thus P (4 out of 400 are defective)

( ) ( ) ( )( )

603.0781.0!42.3

e

2.3008.0400Where;4f008.0,400;4b4

2.3

−=

=

=×=λλ≈=

−

(from table 2 at the end of the text) = 0.178

Cumulative Poisson Distribution Function If X is a rv having Poisson Distribution with parameter ,λ the cumulative Poisson Prob

( ) ( ) ( ) ( )λ===≤=λ= ��==

;kfkXPxXP;xFx

0k

x

0k

For various ( )λλ ;xF,xand has been tabulated in table 2 (of your text book on page 581

to 585) .We use the table 2 as follows.

( ) ( ) ( ) ( )( ) ( )λ−−λ=

−≤−≤===λ;1xF;xF

1xXPxXPxXP;xf

Thus ( ) ( ) ( ) .178.0603.0781.02.3;32.3;42.3;4 =−=−= FFf

Poisson Process There are many situations in which events occur randomly in regular intervals of time. For example in a time period t, let tX be the number of accidents at a busy road junction

in New Delhi; tX be the number of calls received at a telephone exchange; tX be the

number of radio active particles emitted by a radioactive source etc. In all such examples we find tX is a discrete rv which can take non-ve integral values 0,1,2,….. The important

thing to note is that all such random variables have “same” distribution except that the parameter(s) depend on time t. The collection of random variables ( )tX t > 0 is said to constitute a random process. If

each ( )tX has a Poisson Distribution, we say ( )tX is a Poisson process. Now we show

the rvs ( )tX which counts the number of occurrences of a random phenomena in a time

44

period t constitute a Poisson process under suitable assumptions. Suppose in a time period t, a random phenomenon which we call “success” occurs. We let Xt = number of successes in time period t. We assume :

1. In a small time period ,t∆ either no success or one success occurs.

2. The prob of a success in a small time period t∆ is proportional to t∆ i.e. say ( ) tXP t ∆==∆ α1 . ( →α constant of proportionality)

3. The prob of a success during any time period does not depend on what happened prior to that period.

Divide the time period t into n small time periods each of length t∆ . Hence by assumptions above, we note that Xt = no of successes in time period t is a rv having Binomial distribution with parameters n and tp ∆= α . Hence

( ) ( )t,n;xbxXP t ∆α==

( )

��tn.�wherenas�x;f

=∞→→

So we can say that Xt = no of successes in time period t is a rv having Poisson distribution with parameter .tα

Meaning of the proportaratility constant α Since mean of tisX t α=λ , We find α = mean no of successes in unit time.

(Note: For a more rigorous derivation of the distribution of Xt, you may see Meyer, Introductory probability and statistical applications, pages 165-169). Example 19 (Exercise 4.56 of your text) Given that the switch board of a consultant’s office receives on the average 0.6 call per minute, find the probability that

(a) In a given minute there will be at least one call. (b) In a 4-minute interval, there will be at least 3 calls.

45

Solution Xt= no of calls in a t-minute interval is a rv having Poisson distribution with parameter

tt 6.0=α

(a) ( ) ( ) .451.0549.01e10XP11XP 6.011 =−=−==−=≥ −

(b) ( ) ( ) ( ) 430.0570.014.2;2F12XP13XP 44 =−=−=≤−=≥

Example 20 Suppose that Xt, the number of particles emitted in t hours from a radio – active source has a Poisson distribution with parameter 20t. What is the probability that exactly 5 particles are emitted during a 15 minute period? Solution

15 minutes = hour41

Hence 4

1Xif = no of particles emitted in hour41

( )

)2tablefrom(176.0440.0616.0

!55

e!5

2041

e5XP5

5

5

2041

41

=−=

=��

��

�

×== −×−

46

THE GEOMETRIC DISTRIBUTION Suppose there is a random experiment having only two possible outcomes, called ‘success’ and ‘failure’. Assume that the prob of a success in any one ‘trial’ ( ≡ repetition of the experiment) is p and remains the same for all trials. Also assume the trials are independent. The experiment is repeated till a success is got. Let X be the rv that counts the number of trials needed to get the 1st success. Clearly X = x if the first (x-1) trials were failures and the xth trial gave the first success. Hence

( ) ( ) ( ) ( )......2,1xpqpp1p;xgxXP 1x1x ==−=== −−

We say X has a geometric distribution with parameter p (as the respective probabilities form a geometric progression with common ratio q). We can show the mean of this distribution is

p1=µ and the variance is 2

2

pq=σ

(For example suppose a die is rolled till a 6 is got. It is reasonable to expect on an average

we will need 61

61

= rolls as there are 6 nos!)

Example 21 (Exercise 4.60 of your text) An expert hits a target 95% of the time. What is the probability that the expert will miss the target for the first time on the fifteenth shot? Solution Here ‘Success’ means the expert misses the target. Hence ( ) 05.0%5 === SuccessPp . If

X is the rv that counts the no. of shots needed to get ‘a success’, we want

( ) ( ) .05.095.015 1414 ×=×== pqXP

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Example 22 The probability of a successful rocket launching is 0.8. Launching attempts are made till a successful launching has occurred. Find the probability that exactly 6 attempts will be necessary.

Solution ( ) 8.02.0 5 ×

Example 23 X has a geometric distribution with parameter p. show

(a) ( ) ,.........2,11 ==≥ − rqrXP r

(b) ( ) ( )tXPsxtsxP ≥=>+≥ |

Solution

(a) ( ) .qq1pq

.pqrXP 1r1r

rx

1x −−∞

=

− =−

==≥ �

(b) ( ) ( )( ) ( ).1

1

tXPqq

qsXP

tsXPsXtsXP t

s

ts

≥===>

+≥=>+≥ −−+

Application to Queuing Systems

Service facility Customers arrive in a

Depart after service Poisson Fashion There is a service facility. Customers arrive in a random fashion and get service if the server is idle. Else they stand in a Queue and wait to get service. Examples of Queuing systems

1. Cars arriving at a petrol pump to get petrol 2. Men arriving at a Barber’s shop to get hair cut. 3. Ships arriving at a port to deliver goods.

S

48

Questions that one can ask are :

1. At any point of time on an average how many customers are in the system (getting service and waiting to get service)?

2. What is the mean time a customer waits in the system? 3. What proportion of time a server is idle? And so on.

We shall consider only the simplest queueing system where there is only one server. We assume that the population of customers is infinite and that there is no limit on the number of customers that can wait in the queue. We also assume that the customers arrive in a ‘Poission fashion’ at the mean rate of α . This means that tX the number of customers that arrive in a time period t is a rv having

Poisson distribution with parameter tα . We also assume that so long as the service

station is not empty, customers depart in a Poisson fashion at a mean rate of β . This

means, when there is at least one customer, tY , the number of customers that depart

(after getting service) in a time period t is a r.v. having Poisson distribution with parameter tβ (where αβ > ).

Further assumptions are : In a small time interval ,t∆ there will be a single arrival or a

single departure but not both. (Note that by assumptions of Poisson process in a small time interval ,t∆ there can be at most one arrival and at most one departure). Let at time

t, tN be the number of customers in the system. Let ( ) ( ).tpnNP nt == We make another

assumption:

( ) nnn .tastp π∞→π→ is known as the steady state probability distribution of the

number of customers in the system. It can be shown:

( )...,2,1,01

1

=��

��

��

��

−=

−=

nn

n

o

βα

βαπ

βαπ

Thus L = Mean number of customers in the system getting service and waiting to get service)

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αβαπ−

==�∞

=n

n

n.0

qL = Mean no of customers in the queue (waiting to get service)

( ) ( ) βα

αββαπ −=

−=−=�

∞

=Ln n

n

2

1

1

W = mean time a customer spends in the system

ααβL=

−= 1

qW = Mean time a customer spends in the queue.

( ) .1βααββ

α −==−

= WLq

(For a derivation of these results, see Operations Research Vol. 3 by Dr. S. Venkateswaran and Dr. B Singh, EDD Notes of BITS, Pilani). Example 24 (Exercise 4.64 of your text) Trucks arrive at a receiving dock in a Poisson fashion at a mean rate of 2 per hour. The trucks can be unloaded at a mean rate of 3 per hour in a Poisson fashion (so long as the receiving dock is not empty).

(a) What is the average number of trucks being unloaded and waiting to get unloaded?

(b) What is the mean no of trucks in the queue? (c) What is the mean time a truck spends waiting in the queue? (d) What is the prob that there are no trucks waiting to be unloaded? (e) What is the prob that an arriving truck need not wait to get unloaded?

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Solution Here α = arrival rate = 2 per hour

β = departure rate = 3 per hour.

Thus

(a) 223

2 =−

=−

=αβ

αL

(b) ( ) ( ) 34

13222

==−

=αββ

αqL

(c) ( ) hrWq32=

−=

αββα

(d) P (no trucks are waiting to be unloaded)

= (No of trucks in the dock is 0 or 1)

32

32

132

11110 �

��

−+�

��

−=��

��

−+��

��

−=+=

βα

βα

βαππ

95

92

31 =+=

(e) P (arriving truck need not wait)

= P (dock is empty)

= 31

0 =π

Example 25 With reference to example 24, suppose that the cost of keeping a truck in the system is Rs. 15/hour. If it were possible to increase the mean loading rate to 3.5 trucks per hour at a cost of Rs. 12 per hour, would this be worth while?

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Solution In the old scheme, 2,3,2 === Lβα

∴ Mean cost per hour to the dock = 2 x 15 = 30/hr.

In the new scheme 34

L,3,2 ==β=α verify!

∴ Net cost per hour to the dock = .hr/32121534 =+×

Hence it is not worthwhile to go in for the new scheme.

proba

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