12
1/10/2015 1 Circular Permutation Continuation of Ch. 2 CIRCULAR PERMUTATIONS The number of permutations of n distinct objects arranged in a circle is (n – 1)! where one of the objects is considered fixed in its location. Examples: 1. In how many ways can a party of 4 (A,B,C, & D) persons arrange themselves around a circular table? Answer: (4-1)! = 3! = 6 1. ABCD A A---fixed B C D 2. ABDC 3. ACDB 4. ACBD 5. ADBC 6. ADCB In how many ways will A & B sit in adjacent seats? Ans. 4 CIRCULAR PERMUTATIONS 2. In how many ways can a party of 6 (A,B,C, D,E,& F) persons arrange themselves around a circular table? Answer: (6-1)! = 5! = 120 ways In how many ways will: a.) A & B sit in adjacent seats? (2)( 4! ) = 48 b.) A B & C sit in adjacent seats? (3!)(3!)=36 CIRCULAR PERMUTATIONS c. A B & C must not sit in adjacent seats? d. the male (ACE) and the female (BDF) sit alternately? 1. A B C D E F 5. A F C D E B 9. A F E B C D 2. A B E D C F 6. A D C F E B 10. A F E D C B 3. A B C F E D 7. A D C B E F 11. A D E F C B 4. A F C B E D 8. A B E F C D 12. A D E B C F (1)(2!)(3!) = 12 5! – 36 = 84 REMALYN QUINAY-CASEM CHAPTER 3 Conditional Probability and Independence Conditional Probability Multiplication Theorem for Conditional Probability Partition Rule and Bayes’ Theorem Independence, Independent or Repeated Trials

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  • 1/10/2015

    1

    Circular Permutation

    Continuation of Ch. 2

    CIRCULAR PERMUTATIONS

    The number of permutations of n distinct objects arranged in a circle is (n 1)! where one of the objects is considered fixed in its location.

    Examples:1. In how many ways can a party of 4 (A,B,C, & D) persons

    arrange themselves around a circular table?

    Answer: (4-1)! = 3! = 6

    1. ABCD

    A A---fixed

    B

    C

    D

    2. ABDC

    3. ACDB

    4. ACBD

    5. ADBC

    6. ADCB

    In how many ways will A & B sit in adjacent seats? Ans. 4

    CIRCULAR PERMUTATIONS

    2. In how many ways can a party of 6 (A,B,C, D,E,& F)

    persons arrange themselves around a circular table?

    Answer: (6-1)! = 5! = 120 ways

    In how many ways will:a.) A & B sit in adjacent seats?

    (2)( 4! ) = 48

    b.) A B & C sit in adjacent seats?

    (3!)(3!)=36

    CIRCULAR PERMUTATIONS

    c. A B & C must not sit in adjacent seats?

    d. the male (ACE) and the female (BDF) sit alternately?

    1. A B C D E F 5. A F C D E B 9. A F E B C D2. A B E D C F 6. A D C F E B 10. A F E D C B3. A B C F E D 7. A D C B E F 11. A D E F C B4. A F C B E D 8. A B E F C D 12. A D E B C F

    (1)(2!)(3!) = 12

    5! 36 = 84

    REMALYN QUINAY-CASEM

    CHAPTER 3Conditional Probability and Independence

    Conditional ProbabilityMultiplication Theorem for Conditional ProbabilityPartition Rule and Bayes TheoremIndependence, Independent or Repeated Trials

  • 1/10/2015

    2

    Cond

    ition

    al P

    roba

    bilit

    y Suppose that we toss 2 dice, and supposethat each of the 36 possible outcomes isequally likely to occur and hence hasprobability 1/36. Suppose further that weobserve that the first die is a 3. Then,given this information, what is theprobability that the sum of the 2 diceequals 8?

    Cond

    ition

    al P

    roba

    bilit

    y

    In conditional probability problems, the samplespace may be reduced.

    Conditional probability= # of outcomes that satisfy the conditions /

    # of outcomes in the sample space

    Cond

    ition

    al P

    roba

    bilit

    y

    A die is rolled. Find the probability that a 3comes up if it is known that an oddnumber turns up.

    Let T be the event in which a 3 turns upand Q be the event in which an oddnumber turns up.

    Cond

    ition

    al P

    roba

    bilit

    y

    A coin is tossed; then a die is rolled. Findthe probability of obtaining a 6, given thatheads comes up.

    Let S be the event in which a 6 is rolled,and let H be the event in which headscomes up.

    Cond

    ition

    al P

    roba

    bilit

    y

    Two dice were thrown, and a friend tells usthat the numbers that came up weredifferent. Find the probability that the sumof the two numbers was 4.

    Let D be the event in which the two diceshow different numbers, and let F be theevent in which the sum is 4.

    Cond

    ition

    al P

    roba

    bilit

    y

    Two dice are rolled, and a friend tells youthat the first die shows a 6. Find theprobability that the sum of the numbersshowing on the two dice is 7.

    Let S1 be the event in which the first dieshows a 6, and let S2 be the event in whichthe sum is 7.

  • 1/10/2015

    3

    Cond

    ition

    al P

    roba

    bilit

    y

    Assume that 2 cards are drawn in succession and without replacement from a standard deck of 52 cards. Find the probability thata. the second card is the ace of hearts, given that

    the first card was the ace of spades.b. the second card is a king, given that the first

    card was a king.c. the second card is a 7, given that the first card

    was a 6.

    Cond

    ition

    al P

    roba

    bilit

    y

    A coin is flipped twice. Assuming that allsample points are equally likely, what is theprobability that both flips land on heads,given that:(a) the first flip lands on heads?(b) at least one flip lands on heads?

    Cond

    ition

    al P

    roba

    bilit

    y

    Suppose that you hold a ticket in a lotterygame (1-30 numbers) with the numbers 1,14, 15, 20, 23 and 27. You turn on yourtelevision to watch the drawing but all yousee is one number, 15, being drawn whenthe power suddenly goes off in your house.You dont even know whether 15 was thefirst, last, or some in-between draw. What isthe probability that your ticket bears thewinning number combination?

    1 / 29C5 = 0.0000084

    Cond

    ition

    al P

    roba

    bilit

    y

    In the card game bridge, the 52 cards aredealt out equally to 4 players called East,West, North, and South. If North andSouth have a total of 8 spades amongthem, what is the probability that East has3 of the remaining 5 spades?

    (5C3 x 21C10) / 26C13 = 0.339

    Cond

    ition

    al P

    roba

    bilit

    y

    The likelihood of a fatal vehicular crash is affected bynumerous factors. The fatal crashes by speed limit andland use during 2004 are given in the table that follows.

    Suppose a 2004 fatal crash is selected at random.What is the probability that it occurreda. in a rural area?

    Cond

    ition

    al P

    roba

    bilit

    y

    The likelihood of a fatal vehicular crash is affected bynumerous factors. The fatal crashes by speed limit andland use during 2004 are given in the table that follows.

    Suppose a 2004 fatal crash is selected at random.What is the probability that it occurredb. in an area with a speed limit of no more than

    50 mph?

  • 1/10/2015

    4

    Cond

    ition

    al P

    roba

    bilit

    yThe likelihood of a fatal vehicular crash is affected bynumerous factors. The fatal crashes by speed limit andland use during 2004 are given in the table that follows.

    Suppose a 2004 fatal crash is selected at random.What is the probability that it occurredc. in a rural area, given that the speed limit was no

    more than 40 mph?

    Cond

    ition

    al P

    roba

    bilit

    y

    The likelihood of a fatal vehicular crash is affected bynumerous factors. The fatal crashes by speed limit andland use during 2004 are given in the table that follows.

    Suppose a 2004 fatal crash is selected at random.What is the probability that it occurredd. in an urban area, given that the speed limit was

    no more than 40 mph?

    Cond

    ition

    al P

    roba

    bilit

    y

    EXERCISES

    Cond

    ition

    al P

    roba

    bilit

    y

    EXERCISES

    Cond

    ition

    al P

    roba

    bilit

    y

    8. Suppose that two dice were rolled and itwas observed that the sum of the twonumbers was odd. Determine theprobability that the sum was less than 8.

    Cond

    ition

    al P

    roba

    bilit

    y

    9. The numbers of workers, in thousands, in the countryworkforce in 2004 are shown in the table.

    What is the probability that a randomly selected worker is a.male who is at least 65 years of age?b. What is the probability that a randomly selected worker is

    a female?c. What is the probability that a randomly selected worker

    between 16 and 24 years old is a male?d. What is the probability that a randomly selected female

    worker is between 25 and 64 years of age?

  • 1/10/2015

    5

    Mul

    tiplic

    atio

    n Th

    eore

    m

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

    Mul

    tiplic

    atio

    n Th

    eore

    m

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

    Suppose that an urn contains 8 red ballsand 4 white balls. We draw 2 balls from theurn without replacement. What is theprobability that both balls drawn are red?

    P(R1 R2) = P(R1) P(R2 R1)

    P(R1 R2) = (8/12) (7/11)P(R1 R2) = 0.4242

    Mul

    tiplic

    atio

    n Th

    eore

    m Generalized Multiplication RuleM

    ultip

    licat

    ion

    Theo

    rem An ordinary deck of 52 playing cards is

    randomly divided into 4 piles of 13 cardseach. Compute the probability that eachpile has exactly 1 ace.

    Mul

    tiplic

    atio

    n Th

    eore

    m An ordinary deck of 52 playing cards israndomly divided into 4 piles of 13 cardseach. Compute the probability that eachpile has exactly 1 ace.

    Inde

    pend

    ence

    , Ind

    epen

    dent

    or

    Repe

    ated

    Tria

    ls

    Suppose that youre rolling a fair six-sided die. IfA is the event that the die comes up 1 and B isthe event that the die comes up odd, are thesetwo events independent?

    NO!

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

    The conditional probability of A given B equals theprobability of A.

  • 1/10/2015

    6

    Inde

    pend

    ence

    , Ind

    epen

    dent

    or

    Repe

    ated

    Tria

    ls

    Two coins are tossed. Let E1 be the event the firstcoin comes up tails, and let E2 be the event thesecond coin comes up heads. Are E1 and E2independent?

    We have two urns, I and II. Urn I contains 2 redand 3 black balls, whereas urn II contains 3 redand 2 black balls. A ball is drawn at random fromeach urn. What is the probability that both ballsare black?

    independent

    independent (3/5) (2/5) = 6/25 Inde

    pend

    ence

    , Ind

    epen

    dent

    or

    Repe

    ated

    Tria

    ls

    Suppose that two machines 1 and 2 in a factoryare operated independently of each other.Machine 1 will become inoperative during a given8-hour period with a probability of 1/3. Machine2 will become inoperative during the same periodwith a probability of 1/4. Determine theprobability that at least one of the machines willbecome inoperative during the given period.

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

    P(A or B) = 1/3 + 1/4 (1/3)(1/4)

    P(A or B) = 0.5

    Inde

    pend

    ence

    , Ind

    epen

    dent

    or

    Repe

    ated

    Tria

    ls

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

    The conditional probability of A given B equals themarginal probability of A.

    Inde

    pend

    ence

    , Ind

    epen

    dent

    or

    Repe

    ated

    Tria

    lsBob is taking Math, Spanish, and English. Heestimates that his probabilities of receiving As inthese courses are 1/10, 3/10, and 7/10,respectively. If he assumes that the grades can beregarded as independent events, find theprobability that Bob makes

    (a) all As (event A).

    (b) no As (event N).

    (c) exactly two As (event T).

    Inde

    pend

    ence

    , Ind

    epen

    dent

    or

    Repe

    ated

    Tria

    ls

    If two events are independent, does it mean thatthey cant happen at the same time? NO!

    If two events are independent, does it mean thatthey are always mutually exclusive? NO!

    Why? Cite an instance.

    Why? Cite an instance.

    Part

    ition

    Rul

    e an

    d Ba

    yes

    The

    orem

    E = EF + EFC

    PARTITION RULE/ Total Probability

    P(E )= P(EF) + P(EFC )

    P(E )= P(EF)P(F) + P(EFC ) P(FC ) ORP(E )= P(EF)P(F) + P(EFC ) (1 - P(F))

  • 1/10/2015

    7

    Cond

    ition

    al P

    roba

    bilit

    y an

    d PA

    RTIT

    ION

    S

    (60/100) (1/2) + (10/20) (1/2) = 0.46667

    Two boxes contain long bolts and shortbolts. Suppose that one box contains 60long bolts and 40 short bolts, and that theother box contains 10 long bolts and 20short bolts. Suppose that one box isselected at random and a bolt is thenselected at random from that box, what isthe probability that this bolt is long?

    PARTITION RULE/ Total ProbabilityP(E )= P(EF)P(F) + P(EFC ) P(FC )E

    Part

    ition

    Rul

    e an

    d Ba

    yes

    The

    orem

    PARTITION RULE/ Total ProbabilityP(E )= P(EF)P(F) + P(EFC ) P(FC )E

    An insurance company believes that people can be divided intotwo classes: those who are accident prone and those who arenot. The companys statistics show that an accident-proneperson will have an accident at some time within a fixed 1-yearperiod with probability .4, whereas this probability decreases to.2 for a person who is not accident prone. If we assume that 30percent of the population is accident prone, what is theprobability that a new policyholder will have an accident withina year of purchasing a policy?

    (0.4)(0.3) + (0.2)(0.7) = 0.26Suppose that a new policyholder has an accident within a yearof purchasing a policy, what is the probability that he or she isaccident prone? (0.3)(0.4) / 0.26 = 0.4615

    Part

    ition

    Rul

    e an

    d Ba

    yes

    The

    orem

    EXAMPLESuppose that customers have three restaurants to choosefrom in a certain town: R1, R2, R3. Previous data collectionhas shown that these restaurants get 50%, 30% and 20% ofthe customers, respectively. Suppose you also know that 70%of the customers who dine at R1 are satisfied (and 30% arenot), 60% of those who dine at R2 are satisfied, and 50% ofthe customers who dine at R3 are satisfied. What is theprobability that someone who eats at a restaurant in thistown will be satisfied?

    0.5

    0.3

    0.2

    0.70.3

    0.60.4

    0.50.5

    0.5 0.7 =Joint ProbabilitiesFirst R second S

    0.150.180.120.10.1

    0.35

    P(S) = 0.35 + 0.18 + 0.1 = 0.63

    P(E )= P(EF)P(F) + P(EFC ) P(FC )

    Part

    ition

    Rul

    e an

    d Ba

    yes

    The

    orem

    PARTITION RULE/ Total ProbabilityP(E )= P(EF)P(F) + P(EFC ) P(FC )E

    A company buys microchips from three suppliersI, II, andIII. Supplier I has a record of providing microchips that contain10% defectives; Supplier II has a defective rate of 5%; andSupplier III has a defective rate of 2%. Suppose 20%, 35%, and45% of the current supply came from Suppliers I, II, and III,respectively. If a microchip is selected at random from thissupply, what is the probability that it is defective?

    0.20(0.10) + 0.35(0.05) + 0.45(0.02) = 0.0465

    Part

    ition

    Rul

    e an

    d Ba

    yes

    The

    orem

    Bayes Formula

    Part

    ition

    Rul

    e an

    d Ba

    yes

    The

    orem

    Bayes Formula

    Based from the partition ruleP(E )= P(EF)P(F) + P(EFC ) P(FC )

  • 1/10/2015

    8

    Part

    ition

    Rul

    e an

    d Ba

    yes

    The

    orem

    A company buys microchips from three suppliersI, II, andIII. Supplier I has a record of providing microchips that contain10% defectives; Supplier II has a defective rate of 5%; andSupplier III has a defective rate of 2%. Suppose 20%, 35%, and45% of the current supply came from Suppliers I, II, and III,respectively. If a microchip is selected at random from thissupply, what is the probability that it is defective?

    0.20(0.10) + 0.35(0.05) + 0.45(0.02) = 0.0465

    Bayes Formula

    P(A|B) = P(B|A) P(A)P(B|A) P(A) + P(B|AC) P(AC )

    If a randomly selected microchip is defective, what is theprobability that it came from Supplier II?

    (0.05) (0.35) / 0.0465 = 0.376

    Part

    ition

    Rul

    e an

    d Ba

    yes

    The

    orem

    EXAMPLESuppose that customers have three restaurants to choosefrom in a certain town: R1, R2, R3. Previous data collectionhas shown that these restaurants get 50%, 30% and 20% ofthe customers, respectively. Suppose you also know that 70%of the customers who dine at R1 are satisfied (and 30% arenot), 60% of those who dine at R2 are satisfied, and 50% ofthe customers who dine at R3 are satisfied. What is theprobability that someone who eats at a restaurant in thistown will be satisfied?

    0.5

    0.3

    0.2

    0.70.3

    0.60.4

    0.50.5

    0.5 0.7 =Joint ProbabilitiesFirst R second S

    0.150.180.120.10.1

    0.35

    P(S) = 0.35 + 0.18 + 0.1 = 0.63

    Part

    ition

    Rul

    e an

    d Ba

    yes

    The

    orem

    Bayes Formula

    P(A|B) = P(B|A) P(A)P(B|A) P(A) + P(B|AC) P(AC )

    EXAMPLEWhats the chance that a customer ate at R2,given that he or she is satisfied?

    0.5

    0.3

    0.2

    0.70.3

    0.60.4

    0.50.5

    0.5 0.7 =Joint ProbabilitiesFirst R second S

    0.150.180.120.10.1

    0.35

    P(S) = 0.35 + 0.18 + 0.1 = 0.63

    P(R2|S)= 0.18/0.63= 0.286

    or 28.6%

    Part

    ition

    Rul

    e an

    d Ba

    yes

    The

    orem

    Bayes Formula

    P(A|B) = P(B|A) P(A)P(B|A) P(A) + P(B|AC) P(AC )

    EXAMPLEAssuming that the costumer is satisfied, whichrestaurant was he or she most likely to have eaten at?

    0.5

    0.3

    0.2

    0.70.3

    0.60.4

    0.50.5

    0.5 0.7 =Joint ProbabilitiesFirst R second S

    0.150.180.120.10.1

    0.35

    P(S) = 0.35 + 0.18 + 0.1 = 0.63P(R2|S) = 0.286

    P(R1|S)= 0.35/0.63= 0.556

    or 55.6%

    Part

    ition

    Rul

    e an

    d Ba

    yes

    The

    orem

    Bayes Formula

    P(A|B) = P(B|A) P(A)P(B|A) P(A) + P(B|AC) P(AC )

    EXAMPLEAssuming that the costumer is satisfied, whichrestaurant was he or she most likely to have eaten at?

    0.5

    0.3

    0.2

    0.70.3

    0.60.4

    0.50.5

    0.5 0.7 =Joint ProbabilitiesFirst R second S

    0.150.180.120.10.1

    0.35

    P(S) = 0.35 + 0.18 + 0.1 = 0.63P(R2|S) = 0.286

    P(R3|S)= 0.1/0.63= 0.159

    or 15.9%

    P(R1|S) = 0.556at Restaurant 1

    Part

    ition

    Rul

    e an

    d Ba

    yes

    The

    orem

    This makes sense because Restaurant 1 hasthe most customers and gets the highestcostumer satisfaction rating.

    EXAMPLEAssuming that the costumer is satisfied, whichrestaurant was he or she most likely to have eaten at?

    0.5

    0.3

    0.2

    0.70.3

    0.60.4

    0.50.5

    0.5 0.7 =Joint ProbabilitiesFirst R second S

    0.150.180.120.10.1

    0.35

    P(S) = 0.35 + 0.18 + 0.1 = 0.63P(R2|S) = 0.286

    P(R3|S)= 0.1/0.63= 0.159

    or 15.9%

    P(R1|S) = 0.556at Restaurant 1

  • 1/10/2015

    9

    Part

    ition

    Rul

    e an

    d Ba

    yes

    The

    orem

    An industrial company manufactures a certaintype of car in three towns called Farad,Gilbert, and Henry. Of 1000 made in Farad,20% are defective; of 2000 made in Gilbert,10% are defective, and of 3000 made in Henry,5% are defective. You buy a car from a distantdealer. Let D be the event that it is defective, Fthe event that it was made in Farad and so on.Find: (a) P(F|Hc); (b) P(D|Hc);

    (c) P(D); (d) P(F|D).Assume that you are equally likely to have bought any one of the 6000 cars produced. P

    artit

    ion

    Rule

    and

    Ba

    yes

    The

    orem

    (a) P(F|Hc) conditional probability

    Part

    ition

    Rul

    e an

    d Ba

    yes

    The

    orem

    (b) P(D|Hc) conditional probability

    Hc = F G

    60003000

    6000200

    +6000200

    =

    =3000400

    =152

    F G =

    What if F G =

    Part

    ition

    Rul

    e an

    d Ba

    yes

    The

    orem

    (c) P(D) total probability/partition rule

    60003000

    152

    +60003000

    201

    =

    151

    +401

    =

    12011

    =

    Part

    ition

    Rul

    e an

    d Ba

    yes

    The

    orem

    (d) P(F|D) conditional probability

    1201160001000

    51

    =

    =

    12011301

    =114 Pa

    rtiti

    on R

    ule

    and

    Baye

    s T

    heor

    em

    Suppose there is a school having 60% boysand 40% girls as students. The femalestudents wear trousers or skirts in equalnumbers; the boys all wear trousers. Anobserver sees a (random) student from adistance; all the observer can see is that thisstudent is wearing trousers. What is theprobability this student is a girl?

    first stage = sexsecond stage = uniform

  • 1/10/2015

    10

    Part

    ition

    Rul

    e an

    d Ba

    yes

    The

    orem

    A laboratory blood test is 95 percent effective indetecting a certain disease when it is, in fact,present. However, the test also yields a falsepositive result for 1 percent of the healthypersons tested. (That is, if a healthy person istested, then, with probability .01, the test resultwill imply that he or she has the disease.) If .5percent of the population actually has the disease,what is the probability that a person has thedisease given that the test result is positive?

    1st= real health condition2nd = test result Pa

    rtiti

    on R

    ule

    and

    Baye

    s T

    heor

    em

    Consider two urns. The first contains two whiteand seven black balls, and the second containsfive white and six black balls. We flip a fair coinand then draw a ball from the first urn if the coinlands on heads. If tails comes up, draw a ballfrom the second urn. What is the probability thatthe outcome of the toss was heads given that awhite ball was selected?Let W be the event that a white ball is drawn, andlet H be the event that the coin comes up heads.

    ))P(HH|P(W + H)P(H)|P(WH)P(H)|P(W

    =W)|H( ccP

    21

    115

    + 21

    92

    21

    92

    = =

    1986791

    = 67

    22first stage = ?second stage = ?

    Cond

    ition

    al P

    roba

    bilit

    y an

    d In

    depe

    nden

    ce

    EXERCISESCo

    nditi

    onal

    Pro

    babi

    lity

    and

    Inde

    pend

    ence

    EXERCISES

    Cond

    ition

    al P

    roba

    bilit

    y an

    d In

    depe

    nden

    ce

    EXERCISES

    Cond

    ition

    al P

    roba

    bilit

    y an

    d In

    depe

    nden

    ce

    EXERCISES

  • 1/10/2015

    11

    Cond

    ition

    al P

    roba

    bilit

    y an

    d In

    depe

    nden

    ce

    EXERCISES

    Cond

    ition

    al P

    roba

    bilit

    y an

    d In

    depe

    nden

    ce

    EXERCISES

    11. A jar contains 7 red, 6 green, 8 blue, and 4yellow marbles. A marble is chosen atrandom from the jar. After replacing it, asecond marble is chosen. What is theprobability of choosing

    a. a red and then a yellow marble?b. 2 yellow marbles?c. no blue marbles?d. 2 marbles of the same color?e. at least one red marble?

    Cond

    ition

    al P

    roba

    bilit

    y an

    d In

    depe

    nden

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    12. The probability that an archer hits thetarget when it is windy is 0.4; when it is notwindy, her probability of hitting the targetis 0.7. On any shot, the probability of agust of wind is 0.3. Find the probabilitythat:

    (a) On a given shot, there is a gust of windand she hits the target.

    (b) She hits the target with her first shot.

    (c) She hits the target exactly once in two shots. Con

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    13. You enter a chess tournament where yourprobability of winning a game is 0.3 againsthalf the players (call them type 1), 0.4against a quarter of the players (type 2),and 0.5 against the remaining quarter ofthe players (type 3). You play a gameagainst a randomly chosen opponent. Whatis your probability of winning?

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    14. You have a blood test for some rare diseasethat occurs by chance in 1 in every 100 000people. The test is fairly reliable; if youhave the disease, it will correctly say sowith probability 0.95; if you do not havethe disease, the test will wrongly say you dowith probability 0.005. If the test says youdo have the disease, what is the probabilitythat this is a correct diagnosis?

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    15. Two methods, A and B, are available forteaching a certain industrial skill. Thefailure rate is 30% for method A and 10%for method B. Method B is moreexpensive, however, and hence is used only20% of the time. (Method A is used theother 80% of the time.) A worker is taughtthe skill by one of the two methods, but hefails to learn it correctly. What is theprobability that he was taught by usingmethod A?

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    16. John flies frequently and likes to upgradehis seat to first class. He has determinedthat if he checks in for his flight at least 2hours early, the probability that he will getthe upgrade is 0.8; otherwise, theprobability that he will get the upgrade is0.3. With his busy schedule, he checks in atleast 2 hours before his flight only 40% ofthe time. What is the probability that for arandomly selected trip John will be able toupgrade to first class?

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