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Unformatted text preview: MATH 2030 Elementary Probability/ Summer 2009 Midterm exam 1/ June 23, 2009 StudentName: —— DNo.: You have 75 minutes to solve the following problems: Show your complete work.
Except for a calculator and writing utensils no other aides are permitted. The maximal score on this test, including bonus points, is 100. Problem 1 (16 points)
When circuit boards used in the manufacture of compact disc players are tested, the longrun
percentage of defectives is 6%. Let X = the number of defective boards in a random sample of size n=15, so X is binomially distributed with parameters p = 0.06 and n=15.
a) Determine P(X52). b) Determine P(X23). c) Determine P(1sXs4). d) What is the probability that none of the 15 boards is defective? X~ Empaooe)
a) RX 42) =.FC/\/:o) <~ Poem Pan2.) o IS Is ' '9
:12; . .032, i ions0.39
0) 0.06: (,
+69) 00520343
2 g 0.339; + 03123? + 0163/ = 0%23 by f()(;3) : (~PKX‘3) =I‘P(Xé2>
:l~03923 =OO§7I Dc) IPCIéXé Cr) :. 19 /; .O.%2.0343
= (I; 0.39 «(2) _ { 0.062.0f1‘1’2 (. 3 03’3X5 J 0/6}! + 0.0968 + 2100ng '6 = O 603% O. 2 :19} '1 0 It;
not) fun—o) = (é§)Ooe 0.3» Problem 2 (20 points)
Let A and B be events having positive probability. State whether each of the following statements is (i) necessarily true, (ii) necessarily false, or (iii) possibly
true. a) lfA and B are mutually exclusive, then they are independent.
b) If A and B are independent, then they are mutually exclusive.
c) P(A) = P(B) =0.6, and A and B are mutually exclusive. d) P(A) = P(B) =0.6, and A and B are independent. 00 P(A q 3)  o c Mano/((7 emu/rm) => RA (8) = P14 “8) = ole“ P(A)>0
P(B) => 17/4 '3) 3‘ HA) :9) A ahd l8 arc menl
~> 774e, réaiﬁmlmvé «*5 wecersa~237 acetic. fade p—ZMA—évz 1A 5) PC A (,8) = P(A) $08) 5e6awse pf ladepzudwca > O I £V€vz‘(; LlaA/(nl/‘j
POS:\L’7V¢ (“(3117 5? A wad 8 are, Weiz Mo¢ua[(7 2xc(uj?ve. => 7244’. Illa/{.ewec/H‘ PS MeCeern7 {Cdfa. 2k) Pc/z (2/3) : Pc/n + K8) 4%me s KAI) W63) Omaha“? \e‘y¢(u5/Ve) =O,b > I (z :7 7L2 S‘s/«Vlavmeﬁqjl is wzceSJCc/v‘l7 Rafa. 2M) «PL/1mm : PcA).P</)>) Cfmdzpzwdamce) ...> 11,403) ; PC/i) +PC/3)PCAo/3) = 0.610.63 ~O.?6 :089 _._> The “IL/W'c9ers 31¢ CompavéiSQ wc‘ﬂ, PCS) :/ , 3.2. Aug c S, Gitad Mae 54a¥¢m4m¥ f) POLAPLSZV (I‘ve. DJ ['5 £404 M€(€§$6{v;(7 {PL/é , Chad A occh/j cue Vvu/b/uauv «AZXCU/five (\S 304‘ OhO'léIU . Problem 3 (20 points) A computer consulting firm has bids out on three projects. Let A = {project i is awarded}, for
i=1,2,3.. and suppose that P(A1) = 0.22, P(A2)=O.25, P(A3)=O.28, P(A1ﬂA2)=O.11, P(A1ﬂA3)
=0.05, P(A1nAzﬂA3)=O.O1. Compute the following probabilities: More; added! szmg)
I ~— 0.0? 3) P(A2 l A1) b) P(A20A3 I A1) c) P(A2 u A3  A1) d) P(A1ﬂAznA3A1 quuA3) Express in words the probability you have calculated. 01> F(Az//ql) = P(ﬂ2qﬂl); 0” : 03
RA.) 0.22 _ f~0$d$>7(‘1?[7 (020“ project! A: W/(( 9’6
awarded,g>vew {Am’ frOjecv‘l Cm; we” awarded.
5) RA 0A3(Al) ‘ fé/zznﬂanﬂl)
2 £04,)
= 00/  0.049;;
0.22 probaEJ/rQ “nail 570(5) yroj ed; A2 aad A:
w2(( $6. alderman/e” {gout Ppgjec’z (
(40; Size. awwd<6L S I? A? uﬂglﬂ,7 = l’lﬂJ FC/Loﬂlz) ‘ PZ/ImAQJCAMAZn/J) WA, ) 0.1! + 009 ~o_oz
0.22 \
~ (9.68/8 0.22 (01‘ {gaiiz Owe of. {C12
$>¢ awarﬂhd
um: sea», awarde 3) (ﬂuff/d. = ﬂ Amvﬂz 0A3)
FZAI (2/42 U4?) 0?) Péﬂ'L/AZ VA}? 1’ féﬂﬁ (— f6ﬂz7 I PCAI) _/<ﬂ(qA2) " P(ﬂ¢r7ﬂ37 P(ﬂ?oﬂ3) + /(ﬂ,nﬂ20_ﬂz> = 0.92 0.5? P—05a57/5‘(7 “1615 a(( (twee proJ‘ZC‘ZS We
awardad j}:me a¥ 0‘46 (g5 awarﬂ‘éd  Problem 4 (24 points) A qualitycontrol program at a plastic bottle production line involves inspecting finished bottles
for flaws such as microscopic holes. The proportion of bottles that actually have such a flaw is
only 0.0002. If a bottle has a flaw, the probability is 0.995 that it will fail the inspection. lfa
bottle does not have a flaw, the probability is 0.99 that it will pass the inspection. a) If a bottle fails inspection, what is the probability that it has flaw? b) Which of the following is the more correct interpretation of the answer to part (a)? i) Most bottles that fail inspection do not have a flaw.
ii) Most bottles that pass inspection do have a flaw. c) If a bottle passes inspection, what is the probability that it does not have a flaw?
d) Which of the following is the more correct interpretation of the answer to part (c)?
i) Most bottles that fail inspection do have a flaw.
ii) Most bottles that pass inspection do not have a flaw. e) Explain why a small probability in part a) is not a problem, so long as the probability in part
c) is large. Dew04‘6;
A 0. gnga Guns a flaw
B: gag/é Dram roe fur/£0440» PCAC) sagas?
W/BWA) = 0009
div/3 (AC) : 0.0! a) AfP(«7 gavl$f 717?0rem 5 175A I 87 : PUB! x0 I’d/<2)
WZB/A)l’(/4)+P(B(AC).f(/4c) 038900002 :0 (a
OMS0.0002 +0.0103393 40—5; a
.a Cr) 8) Sw/wzZW'WMZ i) ref/£015 111/76 ram/.4
I'M far; 6i)_ Lﬁc) ﬁlo/>07 Bayes/Evie:
WAC/BC) = FCB‘(A‘)f(/<1‘)
F613°(A‘)ﬂ(/4‘) ‘P68C(A)‘PCA) . 0.99.0.‘3938
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wit 0M 07’ ~evew7 200 soH/e; \Laaé are, {(0 wz6{_ Problem 5 (20 points) The prevalence rate in a given community, i.e. the number of people who have a certain
disease, is 0.02. A test is available to diagnose the disease. If a person has the disease, the
probability P1 that the test will produce a positive result is 0.95. If a person does not have the disease, the probability P2 that the test will produce a positive result is 0.01.
a. If a person tests positive, what is the probability that the person actually has the disease?
b. If a man tests negative, what is the probability that he actually has the disease (probability of false negative)?
c. A new diagnostic test is available with performance characteristics P1= 0.98 and P2 =0.10. Calculate the probabilities in b and c for the new test. How would you judge the reliability of
both tests for this disease, and which one is preferable in your opinion and based on your calculations? 01.4401/63 O : I/I FahdOMQ chiem fer:00: Lea;
éhe drmasc. ﬂ: 71L“! Deﬂlcome I‘S fQSr‘L‘Vt Pro): 0.02 ~. fco‘) : 0.32? FLA/D) .— 0.35; H ACID) = 0.0;
J( A/o‘) :0.0(, (PC/IC/OC) : 0.33 l
o 0) ﬂip/J7 807.475! Duh. (13(A .— PCAID>~WDD c
f ) sz/O)‘/’/D)*P(AIO‘)/’ao) ; 0. 699? = MM
033002 4 0.01 0.0.x? 807.25 I pw(¢ ' ﬂat/1‘): PcA‘IO—fﬁ—MWD) c c
N/I‘ID)F(D)+P(ﬂ 10‘)./7(D) = 0031902 _ . 3 0.00!
0.09 0.02 I 003319.39 —————’ x .— proélcw' >)c) for (Cue. new 1/45! . P/ ; P(A/O)  0‘58 ; P(/‘1C/O)‘OOZ f2 .— IVA/0C) .— (mo ; WAC/09.030 => a)’ ﬂow) = 0.3? 0.02 01282? 0.02 (O_/O .0. a? —‘ O. (667
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FfDlﬂ) :O.H967 0‘4(7. Bonus Question (10 points) A friend recently planned a camping trip. He had two flashlights, one that required a single 6V
battery and another that used two sizeD batteries. He had previously packed two 6V and four
size—D batteries in his camper. Suppose the probability that any particular battery works is p
and that batteries work or fail independently of one another. Our friend wants to take just one
flashlight. For what values of p should he take the 6V flashlight? Dzmolc 57 X; Mambo of 6Véaﬂwﬂs
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 Winter '09
 SIKH
 Probability, positive probability, Chad A, Elementary Probability/ Summer

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