Probability 1

PROBALILITYPROBALILITY

THEORYTHEORY

ANDAND

APPLICATIONSAPPLICATIONS

PROBABILITY THEORYPROBABILITY THEORY

Probability Theory is the study of ‘chance’.Probability Theory is the study of ‘chance’. It has vitally important applications in It has vitally important applications in

SciencesSciences Economics Economics

PoliticsPolitics Sports Sports

Life insuranceLife insurance Quality Quality control control Production management Production management

and host of other areas. and host of other areas.

DEFINITIONSDEFINITIONS

ExperimentExperiment::

An act whose outcomes are known in An act whose outcomes are known in advance but is not possible to predict one advance but is not possible to predict one with surety.with surety.

Sample SpaceSample Space (S): (S):

The set of all possible outcomes of an The set of all possible outcomes of an experiment. experiment.

SAMPLE SPACESAMPLE SPACE

SAMPLE SPACESAMPLE SPACE


EventEvent::

Any subset of the sample space.Any subset of the sample space.Equally likely eventsEqually likely events::

Equal chance of happening each of the Equal chance of happening each of the outcome.outcome.

Exhaustive eventsExhaustive events::

The events covering the entire sample The events covering the entire sample space, that is A or B = S. space, that is A or B = S.


Exclusive eventsExclusive events::

Events that do not have any common Events that do not have any common happening.happening.

Sure eventSure event::

An event that is certain to happen.An event that is certain to happen. Impossible eventImpossible event::

An event that is not at all possible to An event that is not at all possible to happen. happen.

MORE ABOUT..MORE ABOUT..

FORMULAEFORMULAE

For any event A,For any event A,

P(A) = n (A) / n (S)P(A) = n (A) / n (S)

And 0 And 0 P(A) P(A) 1. 1.For two events A and B,For two events A and B,

P(A or B) = P(A) + P(B) P(A or B) = P(A) + P(B) P(A and B) P(A and B) If A and B are mutually exclusive events,If A and B are mutually exclusive events,

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

EXAMPLE 1EXAMPLE 1

Find P(A), P(B) and P(A or B).Find P(A), P(B) and P(A or B).

Total number of outcomes = 50Total number of outcomes = 50

68 13

SOLUTIONSOLUTION

n (S) = 50n (S) = 50n (A) = 8 + 6 = 14 n (A) = 8 + 6 = 14 P(A) = 14/50 P(A) = 14/50n (B) = 13 + 6 = 19 n (B) = 13 + 6 = 19 P(B) = 19/50 P(B) = 19/50n (A and B) = 6 n (A and B) = 6 P(A and B) = 6/50 P(A and B) = 6/50P(A or B) = P(A) + P(B) P(A or B) = P(A) + P(B) P(A and B) P(A and B)

= 14/50 + 19/50 = 14/50 + 19/50 6/50 6/50

= 27/50= 27/50

EXAMPLE 2EXAMPLE 2

A fair coin is tossed three times. Find the A fair coin is tossed three times. Find the probability of getting 1 head.probability of getting 1 head.

For the experiment,For the experiment,

Sample Space Sample Space

== S S

= {HHH, HHT, HTH, HTT, THH, THT, = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}TTH, TTT}

i.e., n (S) = 8 i.e., n (S) = 8

Continued..Continued..

Let Event A: getting 1 headLet Event A: getting 1 head

Thus, Thus,

A =A = {HTT, THT, TTH}{HTT, THT, TTH}

i.e., n (A) = 3i.e., n (A) = 3

Hence, probability of getting 1 headHence, probability of getting 1 head

= P(A) = n (A)/n (S)= P(A) = n (A)/n (S)

= 3/8 = 3/8

EXAMPLE 3EXAMPLE 3

A pair of dice is thrown. Find the A pair of dice is thrown. Find the probability of getting (i) a total of 10 (ii) probability of getting (i) a total of 10 (ii) both odd digits (iii) a total that is multiple 3 both odd digits (iii) a total that is multiple 3



S =S = {(1, 1), (1, 2), (1, 3),….., (6, 6)}{(1, 1), (1, 2), (1, 3),….., (6, 6)}

i.e., n (S) = 36 i.e., n (S) = 36

Let event A: total is 10Let event A: total is 10

A = {(4, 6), (5, 5), (6, 4)}A = {(4, 6), (5, 5), (6, 4)}

n (A) = 3n (A) = 3

Thus, P(A) = 3/36 = 1/12Thus, P(A) = 3/36 = 1/12


Let event B: both are odd digitsLet event B: both are odd digits

B = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, B = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)}5), (5, 1), (5, 3), (5, 5)}

n (B) = 9 and P(B) = 9/36 = ¼n (B) = 9 and P(B) = 9/36 = ¼

Let event C: total is a multiple of 3Let event C: total is a multiple of 3

C = {(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, C = {(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3), (6, 6)}6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3), (6, 6)}

n (C) = 12 and P(C) = 12/36 = 1/3n (C) = 12 and P(C) = 12/36 = 1/3

EXAMPLE 4EXAMPLE 4

From a well-shuffled pack of 52 cards, a From a well-shuffled pack of 52 cards, a card is drawn at random. What is the card is drawn at random. What is the probability that it is (i) red (ii) spade (iii) probability that it is (i) red (ii) spade (iii) picture (iv) face (v) heart or king. picture (iv) face (v) heart or king.



S =S = {A{ASS, 2, 2SS, .., K, .., KSS, A, AHH, 2, 2HH, .., K, .., KHH, A, ACC, 2, 2CC, .., , ..,

KKCC, A, ADD, 2, 2DD, .., K, .., KDD}}

i.e., n (S) = 52 i.e., n (S) = 52

Let event A: red card is drawnLet event A: red card is drawn

(Heart and Diamond)(Heart and Diamond)

n (A) = 26 and P(A) = 26/52 = 1/2n (A) = 26 and P(A) = 26/52 = 1/2


Let event B: spade card is drawn Let event B: spade card is drawn

(There are 13 cards of spade)(There are 13 cards of spade)

n (B) = 13 and P(B) = 13/52 = ¼n (B) = 13 and P(B) = 13/52 = ¼

Let event C: picture card is drawn Let event C: picture card is drawn

(There are 16 picture cards)(There are 16 picture cards)

n (C) = 16 and P(B) = 16/52 = 4/13n (C) = 16 and P(B) = 16/52 = 4/13


Let event D: face card is drawn Let event D: face card is drawn

(There are 12 face cards)(There are 12 face cards)

n (D) = 12 and P(D) = 12/52 = 3/13n (D) = 12 and P(D) = 12/52 = 3/13

Let event E: Heart or King card is drawn Let event E: Heart or King card is drawn

(There are 13 hearts, 4 kings; 1 common)(There are 13 hearts, 4 kings; 1 common)

n (E) = 16 and P(E) = 16/52 = 4/13n (E) = 16 and P(E) = 16/52 = 4/13

EXAMPLE 5EXAMPLE 5

Soldier O'Gara, a prison administrator, has Soldier O'Gara, a prison administrator, has been reviewing the prison records on been reviewing the prison records on attempted escapes by inmates. He has attempted escapes by inmates. He has data covering last 45 years that the prison data covering last 45 years that the prison has been open, arranged for seasons. The has been open, arranged for seasons. The data are summarized in the table:data are summarized in the table:

TABLETABLE

EscapesEscapes WinterWinter SpringSpring SummerSummer FallFall

00 33 22 11 00

1-51-5 1515 1010 1111 1212

6-106-10 1515 1212 1111 1616

11-1511-15 55 88 77 77

16-2016-20 33 44 66 55

21-2521-25 22 44 55 332525 22 55 44 22

TotalTotal 4545 4545 4545 4545

EXAMPLE …EXAMPLE …

11 What is the probability that in a year What is the probability that in a year selected at random, the number of selected at random, the number of escapes was between 16-20 during the escapes was between 16-20 during the winter?winter?

22 What is the probability that more than 10 What is the probability that more than 10 escapes were attempted during a escapes were attempted during a randomly chosen summer season?randomly chosen summer season?

SOLUTIONSOLUTION

The probability that in a year selected at The probability that in a year selected at random, the number of escapes was random, the number of escapes was between 16-20 during the winter between 16-20 during the winter

= 3/45 = 1/15.= 3/45 = 1/15. The probability that more than 10 The probability that more than 10

escapes were attempted during a escapes were attempted during a randomly chosen summer season randomly chosen summer season

= 22/45.= 22/45.

READY FOR PRACTICE?READY FOR PRACTICE?

READY FOR PRACTICE?READY FOR PRACTICE?

EXERCISEEXERCISE

EXERCISEEXERCISE

EXERCISEEXERCISE

EXERCISEEXERCISE

EXERCISEEXERCISE

FORMULAEFORMULAE

For exhaustive events A and not A,For exhaustive events A and not A,

P(A) + P (not A) = 1P(A) + P (not A) = 1For independent events A and B,For independent events A and B,

P(A and B) = P(A) P(A and B) = P(A) P(B) P(B)For dependent events A and B,For dependent events A and B,

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

= P(B= P(BA) A) P(A) P(A)

EXAMPLE 1EXAMPLE 1

EXAMPLE 2EXAMPLE 2

Total number of marbles = 5 + 3 + 7 = 15Total number of marbles = 5 + 3 + 7 = 15One marble is selected at a time,One marble is selected at a time,

n (S) = 15n (S) = 15Let event A: the marble is redLet event A: the marble is red

n (A) = 3 and P(A) = 3/15 = 1/5n (A) = 3 and P(A) = 3/15 = 1/5

EXAMPLE 3EXAMPLE 3


Let event B: the marble is greenLet event B: the marble is green n (B) = 5 and P(B) = 5/15 = 1/3n (B) = 5 and P(B) = 5/15 = 1/3

Let event C: the marble is blueLet event C: the marble is blue n (C) = 7 and P(C) = 7/15n (C) = 7 and P(C) = 7/15

Let event D: the marble is not red (not A)Let event D: the marble is not red (not A)

P(D) = 1 – P(A) = 4/5P(D) = 1 – P(A) = 4/5


Let event E: the marble is neither green nor Let event E: the marble is neither green nor blue (A)blue (A)

P(E) = P(A) = 1/5P(E) = P(A) = 1/5

Let event F: the marble is green or redLet event F: the marble is green or red

(not blue)(not blue) P(F) = P (not C) = 1 – P(C) = 8/15P(F) = P (not C) = 1 – P(C) = 8/15

EXAMPLE 3EXAMPLE 3

Number of sample points = n (S) = 36Number of sample points = n (S) = 36

SolutionSolution

A: A: Two 3’sTwo 3’s = {(3, 3)} = {(3, 3)}n (A) = 1 and P(A) = 1/36n (A) = 1 and P(A) = 1/36

B: B: A 5 and a 6A 5 and a 6 = {(5, 6), (6, 5)} = {(5, 6), (6, 5)}n (B) = 2 and P(B) = 1/18n (B) = 2 and P(B) = 1/18

C: C: A 5 or a 6A 5 or a 6 = {(1, 5), (1, 6), (2, 5), (2, 6), = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6), (5, 1), (5, 2), (5, (3, 5), (3, 6), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), 3), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}(6, 5)}n (C) = 18 and P(C) = 1/2n (C) = 18 and P(C) = 1/2

Solution.Solution.

D: D: At least one 6At least one 6 = {(1, 6), (2, 6), (3, 6), (4, = {(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), 6), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}(6, 6)}n (D) = 11 and P(D) = 11/36n (D) = 11 and P(D) = 11/36

E: E: Exactly one 6Exactly one 6 = {(1, 6), (2, 6), (3, 6), (4, = {(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}6), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}n (E) = 10 and P(E) = 5/18n (E) = 10 and P(E) = 5/18

F: F: No 6’sNo 6’s = = not Dnot D P(F) = 1 – P(D) = 1 – 11/36 = 25/36P(F) = 1 – P(D) = 1 – 11/36 = 25/36

Solution..Solution..

G: G: Sum of 7Sum of 7 = {(1, 6), (2, 5), (3, 4), (4, 3), = {(1, 6), (2, 5), (3, 4), (4, 3),(5, 2), (6, 1)}(5, 2), (6, 1)}

n (G) = 6 and P(G) = 1/6n (G) = 6 and P(G) = 1/6H: H: Sum is greater than 8Sum is greater than 8 = sum is either 9 = sum is either 9

or 10 or 11.or 10 or 11.

= {(3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), = {(3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}(6, 3), (6, 4), (6, 5), (6, 6)}

n (H) = 10 and P(H) = 5/18n (H) = 10 and P(H) = 5/18

Solution…Solution…

I: I: Sum of 7Sum of 7 or 11or 11 = {(1, 6), (2, 5), (3, 4), (4, 3),(5, 2), (6, 1), = {(1, 6), (2, 5), (3, 4), (4, 3),(5, 2), (6, 1), (5, 6), (6, 5)}(5, 6), (6, 5)}n (I) = 8 and P(I) = 2/9n (I) = 8 and P(I) = 2/9

J: J: Sum is no more than 8Sum is no more than 8 = Not H = Not HP(J) = 1 - P(H) P(J) = 1 - P(H)

= 1 - 5/18= 1 - 5/18= 13/18= 13/18

EXAMPLE 4EXAMPLE 4

One lottery ticket is drawn at random from a set One lottery ticket is drawn at random from a set of 20 tickets numbered 1, 2, …, 20. Find the of 20 tickets numbered 1, 2, …, 20. Find the probability that the number on the ticket drawn probability that the number on the ticket drawn is divisible by 3 or 5.is divisible by 3 or 5.

Here n (S) = 20.Here n (S) = 20.

Let A: number is divisible is divisible by 3 or 5Let A: number is divisible is divisible by 3 or 5

A = {3, 5, 6, 9, 10, 12, 15, 18, 20}A = {3, 5, 6, 9, 10, 12, 15, 18, 20}

n (A) = 9n (A) = 9

P(A) = 9/20P(A) = 9/20

If the letters of the word THURSDAY be If the letters of the word THURSDAY be arranged at random, what is the probability that arranged at random, what is the probability that the arrangement the arrangement (i) begins with T (i) begins with T (ii) begins with T and ends with U.(ii) begins with T and ends with U.

Two cards are drawn from a well shuffled pack Two cards are drawn from a well shuffled pack of cards. what is the probability that of cards. what is the probability that (i) both are red cards, (i) both are red cards, (ii) both are picture cards , (ii) both are picture cards , (iii) one is a heart card and the other is club, (iii) one is a heart card and the other is club, (iv) both are kings, (iv) both are kings, (v) one of them is an ace card.(v) one of them is an ace card.

In a batch of 400 bolts, 50 bolts are found In a batch of 400 bolts, 50 bolts are found to be defective. A bolt is selected from the to be defective. A bolt is selected from the batch. Find the probability that it is non-batch. Find the probability that it is non-defective.defective.

A bag contains 3 red, 4 blue and 5 green A bag contains 3 red, 4 blue and 5 green balls. Three balls are drawn at random. balls. Three balls are drawn at random. Find the probability that Find the probability that (i) all are blue, (i) all are blue, (ii) all are of the same colour, (ii) all are of the same colour, (iii) no ball is green, (iii) no ball is green, (iv) one ball is red, (iv) one ball is red, (v) two balls are blue.(v) two balls are blue.

In a class of 100 students, 70 failed in Maths, 55 In a class of 100 students, 70 failed in Maths, 55 failed in Physics and 22 failed in both. One failed in Physics and 22 failed in both. One student is selected at random. Find the student is selected at random. Find the probability that he failsprobability that he fails(i) in both subjects,(i) in both subjects,(ii) at least in one subject,(ii) at least in one subject,(iii) neither of the subjects.(iii) neither of the subjects.

A room has 3 sockets. From a collection of a A room has 3 sockets. From a collection of a dozen bulbs of which 4 are defective, 5 are dozen bulbs of which 4 are defective, 5 are selected at random and put in sockets. Find the selected at random and put in sockets. Find the probability that the roomprobability that the room(i) is dark, (i) is dark, (ii) is lighted, (ii) is lighted, (iii) lighted to its maximum.(iii) lighted to its maximum.

FORMULAEFORMULAE

Events A and B are independent if the Events A and B are independent if the happening of A does not affect the happening of A does not affect the happening or non-happening of B and happening or non-happening of B and

P(A and B) = P(A) P(A and B) = P(A) P(B) P(B)Events A and B are dependent if the Events A and B are dependent if the

happening of A does affect the happening happening of A does affect the happening or non-happening of B and or non-happening of B and

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

Two students A and B try to solve a problem. Two students A and B try to solve a problem. Probability of A solving is 3/5 and of B, 1/3. If Probability of A solving is 3/5 and of B, 1/3. If they try independently what is the probability that they try independently what is the probability that the problem is not solved?the problem is not solved?

Given P(A) = 3/5 and P(B) = 1/3Given P(A) = 3/5 and P(B) = 1/3

P(A and B) = P(A and B) = P(A) P(A) P(B) = 3/5 P(B) = 3/5 1/3 = 1/5 1/3 = 1/5

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

= 3/5 + 1/3 – 1/5 = 11/15= 3/5 + 1/3 – 1/5 = 11/15

Thus P (neither A nor B) = 1 – 11/15 = 4/15Thus P (neither A nor B) = 1 – 11/15 = 4/15

EXAMPLE 1EXAMPLE 1

A purse contains 4 silver and 5 gold coins. A purse contains 4 silver and 5 gold coins. Another purse contains 6 silver and 4 gold coins Another purse contains 6 silver and 4 gold coins A purse is selected and a coin is drawn. What is A purse is selected and a coin is drawn. What is the probability that the coin is silver?the probability that the coin is silver?

P (selecting each purse) = P(A) = P(B) = 1/2P (selecting each purse) = P(A) = P(B) = 1/2P (coin is silver) P (coin is silver) = P (silver coin is drawn from A or B)= P (silver coin is drawn from A or B)= P(A) = P(A) P(S P(SA) + P(B) A) + P(B) P(S P(SB) B) = ½ = ½ 4/9 + ½ 4/9 + ½ 6/10 6/10 = 47/90= 47/90

EXAMPLE 2EXAMPLE 2

A purchasing agent has placed rush orders for a A purchasing agent has placed rush orders for a particular raw material with two different suppliers particular raw material with two different suppliers A and B. If neither order arrives in 4 days, the A and B. If neither order arrives in 4 days, the production process must shut down until at least production process must shut down until at least one of the orders arrives. The probability that A one of the orders arrives. The probability that A delivers the material in 4 days is 0.55 and that B delivers the material in 4 days is 0.55 and that B is 0.35. What is the probability that:is 0.35. What is the probability that:(i) both the suppliers deliver material in 4 days? (i) both the suppliers deliver material in 4 days? (ii) at least one supplier delivers the material in 4 (ii) at least one supplier delivers the material in 4 days?days?(iii) the production will be shut?(iii) the production will be shut?

EXAMPLE 3EXAMPLE 3

Given: P(A) = P (A supplies in 4 days) = 0.55Given: P(A) = P (A supplies in 4 days) = 0.55

P(B) = P (B supplies in 4 days) = 0.35 P(B) = P (B supplies in 4 days) = 0.35

Thus, P (not A) = 0.45 and P (not B) = 0.65Thus, P (not A) = 0.45 and P (not B) = 0.65

(i) P (both deliver material in 4 days) (i) P (both deliver material in 4 days)

= 0.55 = 0.55 × 0.35 = 0.1925× 0.35 = 0.1925

(ii) P (at least one delivers the material in 4 (ii) P (at least one delivers the material in 4 days) = P (A but not B) + P (not A but B)days) = P (A but not B) + P (not A but B)

= 0.55 = 0.55 × 0.65 + × 0.65 + 0.45 0.45 × 0.35 × 0.35

= 0.515= 0.515

SolutionSolution

(iii) P (the production will be shut)(iii) P (the production will be shut)

= P (not A and not B)= P (not A and not B)

= 0.45 = 0.45 × 0.65 × 0.65

= 0.2925= 0.2925

There is 29.25% chance that the There is 29.25% chance that the production will be shut because of production will be shut because of unavailability of the raw material.unavailability of the raw material.

SolutionSolution

Small cars get better mileage but are not as safe Small cars get better mileage but are not as safe as big cars. Small cars accounted for 18% of the as big cars. Small cars accounted for 18% of the vehicles on the road but involved in 11898 vehicles on the road but involved in 11898 deaths in accidents. The probability of an deaths in accidents. The probability of an accident involving a small car leading to fatalities accident involving a small car leading to fatalities is 0.128 and the probability of an accident not is 0.128 and the probability of an accident not involving a small car leading to fatalities is 0.05. involving a small car leading to fatalities is 0.05. Suppose you learn an accident involving a Suppose you learn an accident involving a fatality, what is the probability a small car was fatality, what is the probability a small car was involved?involved?

EXAMPLE 4EXAMPLE 4

Given: P(S) = P (small car on road) = 0.18Given: P(S) = P (small car on road) = 0.18

P (not S) = 1 – 0.18 = 0.82 P (not S) = 1 – 0.18 = 0.82

P (fatal when small car) = P (FP (fatal when small car) = P (FS) = 0.128S) = 0.128

P (fatal when not small car) P (fatal when not small car)

= P (F= P (Fnot S) = 0.05.not S) = 0.05.

P (fatal accident with small car) P (fatal accident with small car) = P (S and F) = P (S and F) = 0.18 = 0.18 × 0.128 = 0.02304× 0.128 = 0.02304

P (fatal accident) P (fatal accident) P(F) = P (S and F) + P (not S and F)P(F) = P (S and F) + P (not S and F) = 0.18 = 0.18 × 0.128 + 0.82 × 0.05 × 0.128 + 0.82 × 0.05

= 0.06404 = 0.06404

P (small car involved P (small car involved fatal accident) fatal accident)= P (S= P (SF)F)= P (S and F) / P(F)= P (S and F) / P(F)= 0.02304/0.06404= 0.02304/0.06404= 0.36 = 0.36

A local bank is reviewing its credit card policy A local bank is reviewing its credit card policy with a view toward recalling some of its credit with a view toward recalling some of its credit cards. In the past approximately 5% of the card cards. In the past approximately 5% of the card holders have defaulted and the bank has been holders have defaulted and the bank has been unable to collect the outstanding balance. unable to collect the outstanding balance. Hence, management has established a prior Hence, management has established a prior probability of 0.05 that any particular card holder probability of 0.05 that any particular card holder will default. The bank has further found that the will default. The bank has further found that the probability of missing one or more payments is probability of missing one or more payments is 0.2 for customers who do not default. Of course 0.2 for customers who do not default. Of course the probability of missing one or more payments the probability of missing one or more payments for those who default is 1. for those who default is 1.

EXAMPLE 5EXAMPLE 5

Given that the customer has missed a Given that the customer has missed a payment, compute the probability that the payment, compute the probability that the customer will default. customer will default.

The bank would like to recall its card if the The bank would like to recall its card if the probability that a customer will default is probability that a customer will default is greater than 0.2. Should the bank recall its greater than 0.2. Should the bank recall its card if the customer misses a monthly card if the customer misses a monthly payment? Why or why not?payment? Why or why not?

EXAMPLE 5..EXAMPLE 5..

Tree diagramTree diagram

Given: P(D) = P (customer defaults) = 0.05Given: P(D) = P (customer defaults) = 0.05

P (not D) = 1 – 0.05 = 0.95 P (not D) = 1 – 0.05 = 0.95 Also, P (misses Also, P (misses defaults) = P (M defaults) = P (MD) = 1D) = 1and P (misses and P (misses not default) = P (M not default) = P (Mnot D) = 0.2not D) = 0.2

P (D and M) = P(D) P (D and M) = P(D) × P (M × P (M D) = 0.05 D) = 0.05 × 1 = 0.05× 1 = 0.05

P(M) P(M) = = P(D) P(D) × P (M × P (M D) + P (not D) D) + P (not D) × P (M × P (M not D) not D) = 0.05 = 0.05 × 1 + 0.95 × 0.2× 1 + 0.95 × 0.2= 0.24 = 0.24

P (DP (DM) = 0.05 / 0.24 = 0.21M) = 0.05 / 0.24 = 0.21

EXERCISEEXERCISE

EXERCISEEXERCISE

A game is played using a regular 12-faced fair A game is played using a regular 12-faced fair die, with faces labelled 1 to 12, a coin and a die, with faces labelled 1 to 12, a coin and a simple board with nine squares as shown in the simple board with nine squares as shown in the diagram. Initially the coin is placed on the diagram. Initially the coin is placed on the shaded rectangle, die is rolled and if the shaded rectangle, die is rolled and if the outcome is is prime then coin is moved one outcome is is prime then coin is moved one place to R; otherwise it is towards L. The game place to R; otherwise it is towards L. The game stops when the coin reaches either R or L. Find stops when the coin reaches either R or L. Find the probability that the game the probability that the game (i) ends on 4th move at R,(i) ends on 4th move at R,(ii) ends on 4th move,(ii) ends on 4th move,(iii) ends on 5th move,(iii) ends on 5th move,(iv) takes more than six moves.(iv) takes more than six moves.

In a certain part of the world there are more wet In a certain part of the world there are more wet days than dry days. If a given day is wet, the days than dry days. If a given day is wet, the probability that the next day will also be wet is probability that the next day will also be wet is 0.8. If a given day is dry, the probability that the 0.8. If a given day is dry, the probability that the following day will also be dry is 0.6. following day will also be dry is 0.6. Give that Wednesday of a week is dry, calculate Give that Wednesday of a week is dry, calculate the probability that the probability that (i) Thursday, Friday of the same week are both (i) Thursday, Friday of the same week are both wet,wet,(ii) Friday of that week is wet,(ii) Friday of that week is wet,(iii) In a season, there were 44 matches played (iii) In a season, there were 44 matches played over 3 consecutive days with first and third days over 3 consecutive days with first and third days were dry. How many of these matches expect were dry. How many of these matches expect second day as wet?second day as wet?

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Probability 1

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event b

green n b

event c

cards of spade n b

diamond n

blue n c

event d

event e