Chapter 3 statistics for engineers solution

r

CHAPTER3t o a a a o a a a a a a a a a a a a a a a a a a a a a a a t a a a o a a a a a a a a a a a a a a a a a a o a a a a a a a a a a a a a aa

3.1

Probability

b.

There are five simple events, each corresponding to a classification:

SL, IT, CP, NP, and O

Reasonable probabilities would be probabilities that corespond to the proportions.Thus, the probabilities are:

The probability that the incident oaaurs in a school laboratory is P(SZ) : .06.

There are five simple events, each corresponding to a classification:

BB, TG, GG, S, and G

b. Reasonable probabilities would be probabilities that correspond to the proportions.Thus, the probabilities are:

BBTGGG

^SG

P(BB or G): .28 + .24: .52

Those that support environmentalism in some fashion are the True-blue greens, theGreenback greens, and the Sprouts. The probability is:

P(TG, GG or ^Y)

: .l I + .11 + .26: .48

There are eight simple events, each corresponding to a well class, aquifer, anddetectible level combination:

Simple Event I Probabili

c.

3.3 a.

.28

.11

.11

.2624

c.

d.

3.5

PulB/BL, Pu/BlD, Pu/UBL, PulU/D, Pr/B/BL, PrlBlD, PrlUBL, Pr/U/D

Chapter 3

c.

a.

b. Minitab was used to sort the simple events and the following number of occurrenceswas recorded for each:

b.

PII/B/BL - 57, Pu/B/D - 4l , Pu/U/BL - 15, Pu/U/D - 7,Pr/B/BL - 81, Pr/B/D -22, Pr/U/BL - 0, Pr/U/D - O

The probabilities for the simple events can be found by taking the number ofoccurrenaes for the simple event and dividing by the total number of wells sampled,223.The probabilities are shown below.

Pw/B/BL - .2556, Pu/B/D - .1839 , Pu/U/BL - .0673, Pw/U/D - .0314,Pr/B/BL - .3632, Pv/B/D - .0987, Pr/U/BL - O, Pr/U/D - 0

The probability that a well has a detectible level of MTBE is .1839 + .0314 + .0987 + 0: .3140.

Define the following event:

l: {Beech tree is damaged by fungi}

LOP(A\: '- :.261

188

There would be 3 sample points for this experiment: trrrnk, leaves, and branch. We canconvert the percentages to proportions and use these to estimate the probabilities.

ProbabilitiesTrunkLeavesBranchTotal r.00

From the problem, scientists found that 6 of the 21 families studied had mutations in theSl82 gene on chromosome 14. Thus, a good estimate of this probability is 6/21: .286.

The reliability of this estimate is probably not very high since only 2l families weresampled. To get a more reliable estimate, a larger set of families would be needed.

Let CP: chemical plant and.A/P: non-chemical plant.

P(CP or NP) -- P(CP) + P(NP): .2't + .35 : .56

3.7

3.1 1

Probability

.85

.10

.05

a.

b.

3.9

a.

29

Recall from Exercise 3.1, we have the following probabilities:

Simple Event I Probabili

c.

a.

b. Minitab was used to sort the simple events and the following number of occurrenceswas recorded for each:

b.

Pu/B/BL - 57, Pu/B/D - 4l , Pu/U/BL - 15, Pu/U/D - 7,Pr/B/BL - 81, Pr/B/D -22, Pr/U/BL - 0, Pr/U/D - O

The probabilities for the simple events can be found by taking the number ofoccutrenaes for the simple event and dividing by the total number of wells sampled,223.The probabilities are shown below.

Pw/B/BL - .2556, Pu/B/D - .1839 , Pw/U/BL - .0673, Pw/U/D - .0314,Pr/B/BL - .3632, Pr/B/D - .0987, Pr/U/BL - O, Pr/U/D - 0

The probability that a well has a detectible level of MTBE is .1839 + .0314 + .0987 + 0: .3140.

Define the following event:

l: {Beech tree is damaged by fungi}

LOP(A\: '- =.261

188

There would be 3 sample points for this experiment: trunk, leaves, and branch. We canconvert the percentages to proportions and use these to estimate the probabilities.

ProbabilitiesTrunkLeavesBranchTotal 1.00

From the problem, scientists found that 6 of the 21 families studied had mutations in theSl82 gene on chromosome 14. Thus, a good estimate of this probability is 6/21: .286.

The reliability of this estimate is probably not very high since only 2l families weresampled. To get a more reliable estimate, a larger set of families would be needed.

Let CP: chemical plant and.A/P: non-chemical plant.

P(CP or NP) -- P(CP) + P(NP): .21 + .35 : .56

3-t

3.1 1

Probability

.85

.10

.05

a.

b.

3.9

a.

29

Recall from Exercise 3.1, we have the following probabilities:

Simple Event I Probabili

r

b. Let,SI : school laboratory

P(^92") : I -P(SZ) : | -'06 :'94

3.13 Let PP: procedures and practices.

3.15 a.

P(PP) : | - P(PP): I -.289 : .7tr

The simple events are combinations of whether or not the five chickens pass inspection.Let F : chicken passes inspection with fecal contamination and IZ: chioken passes

inspection without fecal contamination.

The 32 simple events are as follows:

FFFFFFFFFWFFFWFFFFWWFFWFFFFWFWFFWWFFFWWW

FWFFFFWFFIYFWFWFFWFWWFWWFFFIYWFWFWWWFFWWWW

WFFFFWFFFWWFFWFWFFWWWFWFFWFItrFWWFWWFWFWWW

WWFFFWWFFWWWFWFWWFWWWWWFFItrrvwFWWWWWFWWW'YW

b. P(at least one chicken passes inspection with fecal contamination) : I - P(no chickenspass inspection with fecal contamination) : | - P(WWWWW): | - l/25 :24/25 : .96

"1<c. We are told that one in every lO0 chickens pass inspection with fecal contamination in

Exercise 3.6. So, P(n:.01 and P(W):.99. This would make the probabilities of thesimple events listed in part a very different depending on how many lc"s and how manyWsthe simple point contained.

3.17 a. The simple events of the experiment are found by combining the structure type with theactive status of a gas/oil structure. The six simple points would be:

AC, AW, AF,IC,IW,ANdIF

b. Probabilities for the simple events can be found by dividing the number of occurrencesof each simple event by the total number of structures observed (3,400). Theprobabilities are shown in the table below.

Structure

CaissonWell

ProtectorFixed

Platform TotalsActiveInactive

0.148o.176

0.0660.052

o.4260.132

0.6400.360

Totals 0.324 0.1l8 0.558 1.000

P(Active): .64O

P(Well Protector) : .118

Chapter 3

d.

30

e. P(Inactive Caisson) : .148

f. P(Inactive or Fixed Platform) : p(Inactive) + p(Fixed platform) _ p(Both): .36 + .558 - .132: .786

g. P(Caisson"): I -P(Caisson): | - .324- .676

3.19 A simple event for this experiment consists of the outcomes (dry well or oil gusher) for the 6oil wells. Usingthe multiplicative rule, there are atotal of 2 x ) " 2 r 2 x 2-x 2:64 simpleevents' If we let D represent a dry well and, G represent an oil gusher, a simple event will beof the form DDGDGG, where each position coresponds to one of the six wells. If eachsimple event is equally likely, then each simple event has a probability of l/64. Define eventA as the event that at least I oil gusher will be discovered. Of the 64 simple events, only I(DDDDDD) is nor in l. Thus,

3.21

3.23

b.

3.25 a.

Probability

t77 /3,400 - t77 :14451,225

P(A) : | - P(A") : | - 1/64 : 63/64 : .984

I : laser light detects traces of TNTB: object contains no traoes of TNT

A false positive occurs when the laser detects traces of TNT on an object that has no TNTtraces.

We write that as P(A I B):0

a. Define the following events:

{Structure is a fixed platform}{Structure is active}

A:B:

p(BtA_ P(Ar-t B) _. P(B)

P(A I B') - P(A T: B). P(A\

1,447 /3,400 1,447 : '76281,897 /3,400 l,gg7

Define the following events:

l: {Structure is a well protector}.B: {Structure is inactive}

r,225 /3,400

Define the following events:

A: {Cloudy conditions},B: {Unmanned system detects an intruder}

3l

rI

From the table,

P(A):#,, rrur=

P(AnB\:228692

2l+228+226+7 +185 667-_. and692'692

P(B I A\ - P(A a B) - 228 / 692 : -974' P(A) 234/692

b. \D"fine the following events:

C: {Snowy conditions}

From the table,

P(C): -lq692

p(8"):t- p(B)4-#:#, and p(8" .re=#\

P(c l B'): "(9:q,t) =2!61'e2 ='12P(B') 25/692

3.27 Define the following events:

l: {System has high selectivity}.B: {System has high fidelity}

From the problem, P(A) =.72, P(B):.59 and P(A a B) =.33

Thus, P(l 1a1: P(4-2--B)

= '?1 =.ttn' P(B\ .se

3.29 a. Define the following events:

l: {fish is channel catfish}B: {fish is captured 330-350 miles upstream}

From the table,

p(A) =gl#9 =#. rrur: E#q : + . and p(A n q : +P(BlAl-P(AaB)-6/77 = I =.1' P(A\ 60/77 l0

Chapter 3

b. Define the following events;

C: {fish is captured 275-300 miles upstream}D: {fish is small mouth bass}

From the table,

.I: {Intruder}ff: {No intruder}l: {System A sounds alarm},B: {System B sounds alarm}

From the problem:

P(AlD:.e.

P(B I D: .es

P(Alrg:.2

P(B l1v):.l

We know that events A and B are independent. Thus,

P(A a B I D : P(A I D P(B I D: .e(.e5) : .8s5

P(A nB I 1v) : P(A I M P(B I M : .2(.r) : .oz

3.31

J.JJ

b.

Probability

P(c) : r:+]! =#, rro, =W#9 : #,and p(c

^ D : +P(Dlc):'(?2,D = ?.',7J = e

=.225P(C) 40/77 40

Define the following events:

I : {Switch is capable of monitoring the quality of the power running through thedevice)

B : {Switch is wired to monitor the qualify of the power running through thedevice)

We are told in the problem that P(A): .90 and P(B I A): .90.

To find the probability that an Eaton electrical switching device is capable of monitoringpower quality and wired for that purpose, we find:

P(A and B) : P(,4) " P(B I A) : .90 x .90 : .81

Define the following events:

t1

'L*

c.

d. P(Aw B I D: P(Al I) + P(Bl I)- P(A ^ BII): .9 +.95 -.85s :'99s

3.35 Define the following events:

Z- interchange type is TaperedH - traffrc condition is Heavy *A - driver Accepted first lag

a. P(a driver in a tapered merging lane with heavy traffic will accept the first available lag): P(A I both Zand ,F/occur)

We see in the table that 16 drivers accepted and 115 rejected when both T and Hoccurred. So,

P(l Iboth I and f/ occur)' 16+ I 15 131

b. P(a driver in a parallel merging lane will reject the first available lag) : P@l?")

We see in the table that, of the drivers in the parallel type lane, 184 (140 + 44) acceptedand 470 (139 + 331) rejected the first lane. So,

p( A" I T"\ = 47o

= 47o : .72' 184+47O 654

P(driver is in a parallel merging lane given the driver accepts the first available lane inlight traffic)

We see in the table that, of those drivers accepting the first available lane in light traffic,144 of the drivers were in parallel lanes and 67 were in tapered lanes.

_ 144 _r44 :.68267 +144 2II

3.37 Define the following events:

l: {Seed carries single spikelets}.B: {Seed carries paired spikelets}C: {Seed produces ears with single spikelets}D: {Seed produces ears with paired spikelets}

From the problem, P(A) : .4, P(B) : .6, P(Cl A) : .Zg, P(Dl 4 : .7 l, PGI D : .26, andP(DlD:.'74.

a. P(A a C): P(Cl,ayg1: .29(.4):.116

b. p(D): P(A nD)+ P(B ^D): P(Dl4P@)+ P(DlAylq: .71(.4)+ .74(.6): .284 + .444-- .728

34 Chapter 3

3.39 a. Define the following events:

,4: {Failure in earthworks}B: {Failure in earth_retaining structures and excavations}

From the probrem' p(A) : '01 and p(B):.00r. The event fa'ure in either trreearthworks or earth-retaini"g rt u"tures and excavations is A w B-p(Aw B = p(A) + p(B) _ p(A n B)

If A and B are independent events,

P(At.t B) = P(A)P(B) =.01(.001) = .00001

Then, p(A\_i B):.01 +.991 _.00001 :.01099

b' ?x;jElt

c as {Failure in foundations}. The evenr failure in ail 3 design areas is

Ifthe events are independent,

P(An B nc) = p(A)p(B)p(c)= .01(.001)(.0001) = .000000001

a. From the problem, po: p(D n A") and pr : p(D) = p(D n A) + p(D r-t A.) .

Thus, P(A I o1 : P(9 ?-'D - P(D) - P(D n A') = 4 - poP(D) 4-=j-

b. Since I and B cannot occur at the same time,

P(D a A) = p(D n A r_t B") = p1,

P(D a B) = p(D n A" n B) = pr, andp(D) = p(D r-t Aa B") + p(D n A. n B) = pr + pz

Thus, p(l l14:P(2]-A)= 4 andp(,,1o1=p(Dorir _ p,P(D) Pr+pr' u-,-r- pfnl pr+p,

From the problem, Pr = P(D a A) = p(D n A r-t B. ) + p(D a A n B) andPz = P(D r-t B): p(D r-t A. r_t B) + p(D n A r: B)

Since I and, B are independ ent, prp, = p(D o A)p(D n B) = p(D n A n B)P(D) = p(D a An B") + p(D r: A" a B) + p(D n A n B)

:;1;:;;"' + P(D a A n B) + p(D r-t A' n B) + p(D a A n B), p (D n A n B)

3.41

Probability

35

3.43

P(DNA\ NThus, P(l I D) = -F@)- : p, + p, - pf,

P(DrtB\ P2

and P(BlD)=-i(D)== pt+E-ptp, c_

Converting the percentages to probabilities'

P(275- 300) : '52, P(305 -325): '39' and P(330 - 350): '09'

Using BaYes Theorem,

PQ75-300nCC)P(27 s - 300 | cc) = ----FGcl-

b.

P(CCl275-3OO - 300)(cc | 330 - 350)P(330 - 350)

..l

3.45

.775(.52> --

'403 :'403 =.516=ffi .403+ -30o3+-0774 '7807

P( E, r-terror)a. P(E,lerrorl=-jU;"4

P(errorl E)P(E)= *1r.)P(4)

: .or(.30) ,-,==.:. -s-:=:91=.1s8.ot1.ro1*.oH,gToz1sg'003+'006+'01'019

b. P(Erlerror)=ffi

= =' !-\e*orlE')P(E'l =t 'r g,,*

1 niF@ + 4ui' 61 e 6''S + P (error | &) P (4)

_ .006 _ .006 = 316.003 +.006+.01 .019.oX:ol +.03(.20) +.02('50)

Chapter 3

36

c. P(8, I error) - P(E' n error)

P(eror)

P(error I E.)P(E^)P(error I E,)P(E,) + P(error I E)P(E) + P(error | 4)p(4)

.02(.50) .01

.01(.30) +.03(.20) +.02(.50) .003 +.006 +.01 - 'ol :.526.019

d. If there was a serious error, the probability that the error was made by engineer 3 is.526. This probability is higher than for any of the other engineers. Thus engineer #3most likely responsible for the error.

3.47 Define the following events:

,S: {System shuts down}F1: {Hardware failure}F2: {Soltware failure}.F3: {Power failure}

From the exercise, we know:

P(F): .Ol, P(F2): .05, and P(F):.02. Also, P(S I F,) : .73, p(S I F): .tZ, and

P(^S lF3): .88.

The probability that the current shutdown is due to a hardware failure is:

P(4 t^s): o(:.?.") :P(^s) P(sl4)P(4)+P(slF,)p(F,)+p(sl4)p(4)

.73(.01) .0073 - '0073 :.2362.0309.73(.Ol) +.12(.0s) + .88(.02) .0073 + .006 + .0176

The probability that the current shutdown is due to a sofrware failure is:

P(S I F)P(F2)P(.e | 4)P(4) + P(,s I F)P(F)+ P(s I F,)P(n)

.l2(.0s) .006 .006.73(.01)+.12(.0s)+.88(.02) .0073+.006+.0176 .0309

.1942

Probability 3',|

/

The probability that the current shutdown is due to a power failure is:

D/D,o\ _P(F'^,S)_ P(Sl4)P(4)/ \r'3 rol- P(s) - P(sl FJP(rJ-p(sW

.88(.02) .o176.73(.OI) + .12(.0s) +.88(.02) .0073 +.006 + .0176

- '0176 :lssge.0309

3.49 a.

3.5 r

b.

To determine the number of maintenance organization alternatives that exist, we needto consider the number of choices at each of the three echelons. We take the number ofchoices at each echelon and multiply the numbers together to determine that there is

) x 3 x 3 :18 different maintenance organization alternatives

Since four ofthese alternatives are feasible and each has the chance ofbeing selecteddue to the random sarfrpling, then the probability that a feasible alternative is selectedwillbe 4/18: .2222.

For the number 3, there are 3 pa*itions: 3, 2+1, and l+l+1. Now, we must determinehow many ways we can put the color/shape combinations with these partitions.

For the partition 3, there are 4 ways to put the color/shape combinations:

3Y, 3., 36, and 3.1.

For the partition 2+1, there are 16 ways to put the color/shape combinations:

2t+la, 2t+16, 2t+I*, 29+l? ,2a+It, 2a+Ii, 2t+1q, 2a+la,2s+1a, 2^+la, 2i+l*, 2^+Ii,2*+lt, 2*+la, 2*+li, 2*+l*.

For the partition l+l+1, there are 4 ways to put the color/shape combinations:

l9+ lO+ li, 1V+ la+ l*, 1V+ li+ 1*, 1a+ 1i+ l*.

Thus, the total number of colored partitions of the number 3 is 4 + 16 + 4 :24.

For the number 5, there are 7 partitions: 5, 4+1, 3+ 2, 3+l+1, 2+2+1,2+1+l+1, andl+1+l+1+1. Now, we must determine how many ways we can put the color/shapecombinations with these partitions.

For the partition 5, there are 4 ways to put the color/shape combinations:

5Y, 5a, 56, and 5*

Chapter 3

' t+i

b.

For the partition 4+1, there are 16 ways to put the color/shape combinations:

For the partition 3*2,there are l6ways to put the color/shape combinations:

4Y+ la, 49+li,4a+lt, 4a+ 1i,4i+ 1?, 4i+ la,4*+ 1", 4*+ 1.,

3t+2a,3r+2^,3.+2r,3a+2^,3^+21,3^+2a,3&+21,3&+2.,

4v+ l*, 4t+ 19,4a+ l*, 4a+ l),4^+ l.t, 4i+ li,4*+l^, 4*+ 1*.

3a+2*,3e+2t,3a+2&,3a+2a,3^+2T,3i+2i,3*+26,3&+2*.

For the partition 3+l+1, there are 2! ways to put the color/shape combinations:

3?+1a+1.|, 3"+la+l*, 3v+l.t+I.1., 3v+la+1", 3v+li+lt, 3v+l*+l?,3a+lt+li, 3o+la+1.r, 3a+l.l+l*, 3a+14+lv, 3l+fl+lv, 3a+l*+1",3*+la+l'|, 3*+l,i+l*, 3*+la+l.r, 3*+1t+1v, 3*.+14+1v, 3*+l*+lv,3a+1a+1t, 3.l+la+l*, 3i+l.t+l*, 3i+l4+lv, 3t+li+lV, 3.l+l*+1v.

For the partition l+2+2, there are 24 ways to put the color/shape combinations:

la+za+2i, 1t+2a+2*, la+2^+2&, lt+2a+21, la+zi+2r, la+2*+21,la+2a+2^, la+za+Z*, la+2^+2*, la+2a+2?, Ia+2^+2f, l)+2*+2t,l*+2a+2^, l*+2a+2*, I*+2^+2*, 1&+Z|+2r, l*+2t+2r, l*+2&+21,l^+2a+2i, l^+2a+2*, l6+2d+2*, l^+2a+21, l^+2i+2r, l^+2&+2a.

For the partition 2+1+l+l,there are l6 ways to put the color/shape combinations:

2v+la+16*1.?., 2"+1a+li+lv, 2v+1a+lv+l*, 2v+lr+1.l+1*,2t+la+1.t+l*, 2t+la+li+lr, 2r+ll)+1v*1.?., 2a+1v+1.l+1.r,2*+1.+1lt+l*, 2&+la+li+lY, 2*+1 a+l ?+1jr, 2*+l v+14+1rF,2A+1 r +1 l|+l *, zi+I a+l.l+l v, 2 .t+l a+1?+1*, 2a+l v+l i+l *,

For the partition l+l+l+l+1, there are 0 ways to put the color/shape combinations.

Thus the total number of colored partitions of the number 5 is

4 + 16+ 16 + 24 + 24+ 16 + 0 : 100.

There are a total of 4 x 4: 16 metal-support combinations possible'3.53

3.5 5

Probability

For each of the metals, there are

the four metals.

Pno = 4l', : 4' 3' 2' l :24differentorderingsof

" (4-4)t

To find the number of different responses to the questionnaires, the engineers mustevaluate 2 building parameters for each of 3 parts. Using the multiplicative rule, there

39

b.

are 2 x 3 : 6 parameter-part combinations to evaluate. Each of the 6 parameter-partcombinations is to be rated with one of 3 responses. Again, using the multiplicativerule, there are a total of

3x3x3x3x3x3:36:729

different possibte responses to the questionnaire.

The engineers want to select the 3 parameter-part combinations with the highest ratingsfrom the 6 possible combinations and rank them in order. Using the permutation rule,there are a total of

p6_ 6!'r - 16-3;9-

6x5x4x3x2xl3x2xl =l2O

3.57 a.

b.

There are 2 x 2 x 6 x 7 : l6Smachining conditions possible.

The probability of detecting the flaw is 8/168 : .0476.

If we limit drill size to .25 inch and drill speed to 2,500 rpm, there are only 2 x 7 : 14

different machining conditions possible. Since the study included two of thosepossibilities, the probability of detecting the flaw in the study is2/14: ,L429.

,.+3.59 i*.Using the partitions rule,

l6! :63.063,0004t4l4t4l 4.3.2.1 4.3.2.1

3.61 a. To find the probability of the dealer drawing blackjack, we must find the total numberof ways to get blackjack and the total number of ways to draw two cards. The totalnumber of ways to draw 2 cards from a deck of 52 cards is found using thecombination rule and is

( sz\ szt s2t 52x51I l: =-=-:132612 ) 2l(s2-2)l 2ts0t 2xlx5ox49

To find the total number of ways to get blackjack, we must multiply the number ofways to get I ace and the number of ways to get I card with a value of 10. The numberof ways to get 1 ace is found using the combination rule and is

(+) 41 4t 4x3x2xtI t_ -A

---

(1./ l!(4-l)! l!3! lx3x2xl

The number of ways to get I card with a value of l0 (there are 16 cards with a value of10) is found using the combination rule and is

c.

(to\ 16! 16!I t-

I r ./- r(r6-r)! Ms!l6x 15 x'..x I

lx15x14x...xl =16

Chapter 3

L

b.

The total number of ways to draw blackjack is 4 x 16: 64.

The probability of drawing blackjack is 64/7326.

Define the following events:

B: {Player draws Blackjack}D: {Dealer draws Blackjack}

The event we want to find is P(Bn D") .

We know that P(B a D') = P(B)P(D' I B) .

Frompart a, P(B)= 64

1326

Now we will find P(D]B). If the player has already been dealt blackjack, then there are

only 50 cards left from which the dealer draws 2. Thus, the total number of ways ofselecting 2 cards from 50 is

/so\ so! so! s0.49l--l:\z ) 2t(so-2\t 2t48! 2

To find the total number of ways the dealer can get blackjack given the player has

already been dealt blackjack, we must multiply the number of ways to get I ace given

the player has I ace andlhe number of ways to get I card with a value of 10 given the

ptayer -has

1 card with a value of 10. The number of ways to get I ace given the player

has I ace is found using the combination rule and is

(l)3!3^I t:-Il] l!(3-l)! 1t2l

The number of ways to get I card with a value of 10 given the player has I card with avalue of l0 is found using the combination rule and is

(ts) rs! rsI l=

I r, 1!(ls-l)! l!14!

The total number of ways to draw blac(ack given the player has drawn blackjack is

3 x 15:45.

as 45 I 180P(DlB'):

= andP(D' lB)=t-P(DlB):l-

- =:-:-1225*'-^'- '-' 1225 1225

Finally, P(B n D') = P(B)P(D' I B) : # #= m=.o465

41Probability

3.63 Define the following events:

l: {CD i is not defective} P(A,) = .99

If we assume that the claim is true and all the CD's are independent, then the probability thatnone of the CD's are defective is:

P(A, n A, r-t A, r-t Aq) = P(4)P(A)P(4)P(A,) = .994 = .961

Thus, the probability that at least I CD is defective is the probability of the complement ofthe above event or I - .961 : .039. If the claim is true, it would be very unlikely that at leastone of the next 4 CD's manufactured will be defective (probability .039). We would inferthat the probability that a CD is defective is greater than I in 100.

3.65 a. Define the following events:

l, : { lst antiaircraft shell strikes within 30 feet of target}

Ar: {2nd antiaircraft shell strikes within 30 feet of target}Ar: {3rd antiaircraft shell strikes within 30 feet of target}

From the text, P(A,): p: .45 for i: 1,2,3

P(All 3 shells miss their targets): P(Ai n Ai r-t Ai) = P(AI)P("4;)P(A:): u - P(A)ltr - P(A)III - P(4)1: (l - .4sxr - .4sxl - .4s): .166

Assume the simple events are independent. Since this is not a very small probability,there is no evidence to conclude that in battle conditions, p differs from .45.

Define events Aa, \, ... , 4,oas in part a for shells 4 through 10.

P(All l0 miss their targets) : P(Ai r\ A;." ^ lio) : P(Ai)P(Ai)"'P(Aio)

(l - .45)to:.0025

Since this is a very small probability, there is evidence to conclude that in battleconditions,p differs from .45.

The experiment consists of selecting a rule and recording its type. The simple eventsare the list of all possible types.

Batch scheduling, JES queue space, C-to-C links,Hardware errors, SMF management, Quiesce and IPL,Performance, and Background monitor.

Chapter 3

b.

3.73 a.

b. Since there are 548 different rules and differing numbers of rules per simple event, we

will assign probabilities to each simple event corresponding to the number of rules per

simple event divided by the total number of rules, 548' Thus,

P(Batch schedule) : 139/ 548,P(JES): rO4/548,

P(C-to-C links) : 68/548,P(Hardware errors) : 87 / 548,

P(SMF management) : 25/548,P(Quiesce and IPL) : 52/548,

P(Performan ce) : 4l / 548, ffidP(Background monitor) : 32/ 5 48.

Define event A as {C-to-C link or Hardware error}

P(A): 68/548 + 87/548: 15s/s48

Define event -B as {Not performance}. Event ,B contains all the simple events exceptperformance. Thus,

P(B) : 139/548 + lO4/548+ 68/548 + 87/s48 + 2s/s48 + s2/548 + 32/548:507/548

Let Abe the event that a dragonfly species inhabits a dragonfly hotspot. ThenP(A):.e2.

Let B be the event that a dragonfly species inhabits a dragonfly hotspot. ThenP(B): .e2.

Since 1.00 or lOO%o of all butterfly species inhabit bird hotspots, then the butterflyhotspots have to be part ofthe bird hotspots. Thus, all butterfly hotspots are contained

in bird hotspots.

The total number of tablets selected was 6 per lot x 30 lots : 180. Eight measurements

were taken on each tablet, so the total number of measurements is:

180 x 8 :144O

An average measurement was computed for each lot (30) at each time period (8). Thetotal number of averages computed is:

30 x 8 :240

Define the following events:

l, : {Shuttle mission i does not result in a critical-item failure}

P(A,'): | -l/63:62/63

c.

d.

3.75

3.77 a.

3.79 a.

Probability

b.

b.

43

b.

Then the event that none of the 8 shuttle flights results in a critical-item failure can be

written as A, n A, n A, r-t A" r:4 n Au r-t.\ n At. The event that at least one of the 8

shuttle flights results in a critical-item failure is:

(Ar n Az rt A, n Aq n 4 n Au a'4, r\ 4)'

Thus,

P(At n A, r-t A, n Ao a,4, n Au n A, ^ 4)"

: | - P(4 r\ 4 ^ At n Ao r: A, n'4u n A, a A")

: 1 - P (4) P (A) P (4) P (A4\ P ('4) P (.4A P Q4r) P (4): | - (62/ 63)(62 / 63)(62 / 63)(62 / 63)(62 / 63)(62 / 63)(62 / 63)(62 / 63): I - .880 : .l2O (assuming the missions are independent of each other)

Using the same logic as above, the probability that at least one of the 40 shuttlemissions will result in a critical-item failure is:

P(A, r t A, r-t A, r:... n Aoo)o

: | - P('4\ a'4t n At r: "' a Aoo)

: | - P(4)P(A)P(4)... P(A+o)

: l-(62/63)ao:l-.527:.473

Define the following events:

l: { lst system breaks down}B: {2nd system breaks down}

From the problem, P(A): .2 and P(81,4): .3. The event the system is not working isAnB.

P(A n B) = P(B I A)P(A) : .2(.3) = .06

The event the system is working is the complement of event A n B .

P(A n B)' :l - P(A^.8) = I -.O6 =.94

The probability that a single oil well prospect will result in no more than 100,000barrels of oil is;

P(0) + P(50,000) + P(I00,000): .60 + .10 + .15 : .85

The probability that a single oil well prospect will strike oil is the complement of theevent that a single oil well prospect will not strike oil and is:

1-P(0):1-.60:.40

Chapter 3

3.81 a.

3.83

b.

b.

44

If two wells are drilled, the total number of outcomes is 5 x 5 :25 using themultiplicative rule. The simple events are (in thousands):

(0,0) (0,50) (0,100) (0,500) (0, 1000)(s0,0) (s0, s0) (50, 100) (s0, s00) (s0,1,000)(100,0) (100,50) (100, 100) (100,500) (100, 1,000)(500,0) (500,50) (500, 100) (500, 500) (500, 1,000)(1,000,0) (1,000,50) (1,000, 100) (1,000,500) (t,000, 1,000)

d. We are given in the problem thatP(0):.6, p(50): .1, p(I00): .15, p(500): .1, andP(1,000) : .05. The event (0, 0) is really the event 0 n 0 .

Since the outcomes of the two wells are independent,

P(0, 0) : P(0n 0): p(0)p(0): .6(.6) : .36

Similarly, P(0, 50) : P(0)P(50) : .6(.1): .06

The probabilities to all the simple events are listed below:

P(0, 0): .6(.6): .36P(0, 100): .6(.15):.09

P(0' 1,000): .6(.05): .03P(s0, s0): .l(.1): .01

P(sO, 500): .l(.1) - .01P(100, 0): .l s(.6): .0e

P(100, 100):.15(.1s) : .022sP(l00, 1,000) : .15(.05) : .0075

P(s00, s0): .l(.1): .01P(500, s00): .l(.1): .01P(1,000, 0): .0s(.6): .03

P(1,000, 100): .05(.15): .007sP(I,000, 1,000):.05(.05) : .0025

P(0, 50): .6(1):.06P(0, s00): .6(.1):.06P(sO, 0): .l(.6): .06

P(sO, 100): .1(.15): .015P(50, 1,000): .1(.05): .005P(1,000, 50) : .15(.1) : .015

P(1,000, 500): .15(.1): .015P(s00, 0): .1(.6):.06

P(s00, 100): .l(.ls): .01sP(500, 1,000) : .l(.05) : .005P(1,000, 50) : .05(.1) : .005

P(1,000, 500): .05(.1): .005

e. Define the event:

A: {At least 1 of the 2 oil prospects strikes oil}

Then A'is the event that neither of the two oil prospects strikes oil.

P(A') : P(0, 0) : .36

Thus,P(l) : | - P(A"): I - .36: .64

Probability 45

3.85 There are five suppliers and a company will choose at least 2 suppliers. Therefore, we mustfind the number of ways to select 2,3,4, and 5 suppliers from the 5 suppliers. The totalnumber of options is:

5! 5! 5! 5!+-+zt(s-2)t 3!(s-3)! 4!(s - 4)t s!(s-s)!

5x4x3x2xl 5x4x3xZxl2xlx3x2xI 3x2xlx2xl

5x4x3x2xl 5x4x3x2xl : l0+10+5+l=26

3.87

4x3x2xlxl 5x4x3x2xlxl

Decide on a starting point on the random number table. Then take the first z numbersreading down, and this would be the sample. Group the digits on the random numbertable into groups of 7 (for part b) or groups of 4 (for part c). Eliminate any duplicatesand numbers that begin with zero sinoe they are not valid telephone numbers.

Starting in Row 6, column 5, take the first l0 seven-digit numbers reading down. Thetelephone numbers are:

277-5653988-7231t88-7620174-s318530-6059709-9779496-2669889-7433482-3752772-3313

Starting in Row 10, column "l,take the first 5 four-digit numbers reading down. The 5telephone numbers are:

373-3886373-s686373-1866373-3632373-6768

There are 3 contracts to be awarded. Since no company can be awarded more than onecontract, there are five possible companies to which to award the first contract, onlyfour companies to which to award the second contract, and three companies to which toaward the third contraat. The total number of ways to award the contracts is foundusing the permutation of five companies taken 3 at a time or

[;) . [;). [;). [;)

=

b.

c.

3.89

_ <ta5

" - (5-3)!-5x4x3x2xl

2xl =60

46 Chapter 3

b. If company 2 is awarded the first contract, then there are only 4 compalies to which toaward the second contract and only 3 companies to which to award the third contract.Thus, the total number of ways to award contracts so company 2 receives the firstcontract is the permutation of 4 companies taken2 at a time or

4l 4x3xZxl- la

2xlP;:

(4 -2)l

3.91 a.

Probability

Company 2 could also be awarded the second or third contract. There are also 12 wayscompany 2 could receive the second contract and 12 ways company 2 could receive tlethird contract. Thus, the total number of ways company2 could be awarded a contraatis 12 + 12 + 12:36.

P(Company 2 awarded a contract) :36/60: .6

There are 6 ways companies 4 and 5 can be awarded 2 of the contracts. One way is forcompany 4 to receive the l"t and company 5 the 2nd. The remaining ways are:company 4 the l"t and company 5 the 3d., company 4 the 2"1ana

"impuny 5 the 3'd,

aompany 5 the I't,and company 4theZ"d,"o-pu.ty 5 the l't and "ompany

+tn"3d,'andcompany 5 the 2"d and company 4 the 3'd. noiany of the above mentionld ways thatcompanies 4 and 5 can receive contracts, there are 3 companies left from whicir toselect I to fill in the remaining contract. Thus, the total number of ways companies 4and 5 can receive contracts is 6 x 3 : 18.

P(Company 4 and company 5 awarded contracts) -- l g/60 : .3

A flush is 5 cards of the same suit. The total number of ways to get a flush in one suitis a combination of 13 cards taken 5 at a time or

/r:\ r ?rIt:Is] 5!(t3-s)!

Since there are four suits, the total number of ways to draw a flush is ae2g7): 514g.The total number of ways to draw 5 cards from a deck of 52 is:

(sz.r \)tIt J: "(t-t

=2'5e8'e6o

Thus, P(l) : 5148/2,598,960 : .0019808

47

b. A straight is 5 cards in sequence, regardless of suit. The total number of ways to get astraight starting with an ace is:

f)[)(r)(r)fi)(combination of 4 aces taken I at a time times a combination of 4 twos taken one at atime, etc.)

4l 4l 4l 4l 4l1(4-tI " l(4{X x

1(4 {X * 14+ -Ur* l(4 -lX

ls: lO24

Flowever, astraightcanstartwith an ace,2,3,... ,10. The total number ofways todraw a straight is 1024(10) : lO,24O.

P(B) : 10,240/2,598,960 : .00394

c. There are 10 ways to get a straight flush in one suit (straight starting with an ace,2,3,. .. , l0). Since there are four suits, the total number of ways to draw a straight flush isl0(4):40.

P(An8):40/2,598,960:.0000154 I

3.93 Suppose we define the following event:

.E: {Error produced when dividing}

From the problem, we know that P(E): I / 9,000,000,000

The probability of no error produced when dividing is:

P(E) : I - P(E) : I -l / 9,000,000,000 : 8,999,999,999 / 9,000,000,000: .999999999 = 1.0000

Suppose we want to find the probability of no errors in 2 divisions (assuming each division isindependent):

P(E n E): .999999999(.999999999): .999999999 n: 1.0000

Thus, in general, the probability of no errors in & divisions would be:

P(t n E r: E n ...- E) : P(Ef : [8,999,999,999 / 9,000,000,000]tfr times

Chapter 348

Suppose a user ran a program that performed I billion divisions. The probability of no errorsin these 1 billion divisions would be:

P(f ; t'ooo'ooo'000 : [8,999,ggg,g99 / 9,000,000,0001 r'ooo'otto'000 :

. 904 8

Thus, the probability of at least I error in I billion divisions would be

l-P(A)t'oo0'000'000:1-18,999,ggg,gg9l9,000'000'0001r'ooo'ooo'ooo:1-'9048:'0852

For a heavy SAS user, this flawed chip would be a problem because the above probability isnot that small.

Probability 49