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Unformatted text preview: ZOOB—ASL
M 8: S HONG KONG EXAMINATIONS AND ASSESSMENT AUTHORITY HONG KONG ADVANCED LEVEL EXAMINATION 2008 MATHEMATICS AND STATISTICS ASLEVEL 8.30 am —11.30 am (3 hours) This paper must be answered in English 1. This paper consists of Section A and Section B. 2. Answer ALL questions in Section A, using the AL(E) answer book. 3. Answer any FOUR questions in Section B, using the AL(C) answer book.
4. Unless otherwise speciﬁed, all working must be clearly shown. 5. Unless otherwise speciﬁed, numerical answers Should be either exact or given to 4 decimal places. ZOOSASM & 8—1. 14 Section A (40 marks)
Answer ALL questions in this section.
Write your answers in the AL(E) answer book. 1 . . . 3
m ascendmg powers of x 13 1+ ——x + bxz +   . where l+rrx 2 1. (a) The binomial expansion of
a and b are constants. (i) Find the values of a and E). (ii) State the range of values of x for which the expansion is valid. 1 (b) Using (a) with x = 3—0 _, ﬁnd an approximate value of V10 .
(7 marks)
2
2. Suppose y3—uy=l and 24:23: .
. dy .
(a) Ford d— 111 terms of u and y.
u
. du .
(b) Fmd — m terms of x.
dx
. dy .
(c) Fmd a; In terms of x and y.
(7 marks) 3. The rate of change of concentration of a drug in the blood of a patient can be modelled by Eli=5.3 —1———; 1.2.941“ ,
dt r+2 r+5 where x is the concentration measured in mg/L and r is the time measured in hours after the
patient has taken the drug. It is given that x = 0 when t = O . (a) Find at in terms of r. (b) Find the concentration of the drug after a long time.
(6 marks) ZOOSASM a 8—2 I 15 4. A and B are two events. A' and B’ are the complementary events of A and B respectively. Suppose P(A)=% , P(AUB)=210, P(AB)=% and P(B)=k ,where 0<k<1. (a) Using P(AB) ,express P(AmB) interms of k.
(in) Find the value of k. (c) Find P(A’mB). (d) Are the two events A' and B' mutually exclusive? Explain your answer.
(7 marks) 5. The amount of money spent by a randomly selected customer of a jewellery shop is assumed to be
normally distributed with a mean of $ ,u and a standard deviation of $6 000 . Suppose 24.2% of the customers spend more than $30 000 in the shop. (3) Find the value of n . (b) It is given that Mrs. Chan spends less than $30 000 in the shop. Find the probability that she spends more than $16 500 .
(6 marks) 6. A test is taken by a class of 18 students. The marks are as follows: 55 32 74 70 91 75 79 89 68
79 59 72 79 73 60 7182 k where k is J ane’s mark. It is known that the mean mark of the class is the same irreSpective of including or excluding Jane’s. (3.) Find the value. of k. (b) If 3 student marks are selected randomly from the set of the 13 student marks. find the
probability that exactly 1 of them is the mode of the set of the 18 student marks. (c) A student mark is classiﬁed as an outlier if it lies outside the interval (,u — 20'. ,u + 20') ,
where ,u is the mean and o" is the standard deviation of the set of marks. (1') Find all the outlier(s) of the set of the 18 student marks. (ii) In order to assess the students’ performance in the test, all outliers are removed from the set. Describe the change in the median and the standard deviation of the student marks
due to such removal. (7 marks) ZOOS—ASM & 3—3 16 Section B (60 marks) Answer any FOUR questions in this section. Each question carries 15 marks.
Write your answers in the AL(C) answer book. 3x—2
x+2 7. Deﬁne. f(x) = for all x :2 ——2 . Let C be the curve y = f(x) . (a) Find the equations of the horizontal asymptote(s) and vertical asymptote(s) to C .
(2 marks) (in) Sketch C and its asymptotes. Indicate the point(s) where the curve cuts the axes.
(3 marks) (c) PM, It) is a point on C in the ﬁrst quadrant Let L1 and L2 be reSpectively the tangent
and normal to C at P. (i) Show that the equation of L1 is
8x—(h+2)2y+3h2 —4h—4 =0. (ii) If L1 passes through the origin, find
(1) the equation of L2, (2) the area of the region bounded by C , L2 and. the xaxis.
(10 marks) BOOBASM & 8—4 17 3. A biologist studied the population of fruit ﬂy A under limited food supply. Let t be the number of
days since the beginning of the experiment and NO) be the number of fruit ﬂy A at time t. The biologist modelled the rate of change of the number of fruit fly A by 20 "20* N’(t) = where h and k are positive constants. 20
N’G‘) (a) (i) Express ln[ —1] as a linear function of z‘. (ii) It is given that the intercepts on the vertical axis and the horizontal axis of the graph of the linear function in (i) are 1.5 and 7.6 respectively. Find the values of h and k.
(4 marks) (b) Take 11 = 4.5 ,, k = 0.2 and assume that N(O) = 50 . (i) Let v=h+ek’ ,ﬁnd —. Hence, or otherwise, ﬁnd NU) . (ii) The population of fruit ﬂy B can be modelled by M(z)=21[z+%e“’“]+b _, where b is a constant. It is known that M(20) = N(20) . (1) Find the value of e. (2) The biologist claims that the number of fruit ﬂy A will be smaller than that of fruit
fly B for z‘> 20 . Do you agree? Explain your answer. [Hint Consider the difference between the rates of change of the two populations]
(11 marks) 2003ASM 36 3—5 13 The rate of change of yearly average temperature of a city is predicted to be
dx 1 l 2
— = — 1+: r 2 0 ,
dr 40 ( ) where x is the temperature measured in “C and t is the time measured in years. It is given that
3: =22 when i=0. (a) (i) Using the trapezoidal rule with 4 subintervals, estimate the increase of temperature
from r=0 to 1:10. (ii) Determine whether this estimate is an ovenestimate or an underestimate.
(4 marks) (b) It is known that the electricity consumption W(x) , in appropriate units, depends on the
yearly average temperature 3: and is given by won = 100011302 — 6301nx +1960 (x 2 22) . (i) If Wag) = 968 ,ﬁnd all possible value(s) of x0 . (ii) Find the range of values of x while W’(x) < 0 . (iii) Find the rate of change of electricity consumption at r = 0 . (iv) Using (a), estimate the electricity consumption at 2* =10 . Determine and explain whether the actual electricity consumption is larger than or smaller than this estimate.
(11 marks) ZOOSAS—M & S—6 19 10. Assume that the number of visitors arriving at each counter in an irrnnigration hall is independent and follows a Poisson distribution with a mean of 3 .9 visitors per minute. A counter is classiﬁed as
busy if at least 4 Visitors arriving at it in one minute. (a) Find the probability that a counter is busy in a certain minute.
(3 marks) (b) An ofﬁcer checks 4 counters in a certain minute. Find the probability that at least one busy
counter is found. (2 marks) (c) If 10 counters are open, ﬁnd the probability that more than ’7 of them are busy in a certain
minute.
(3 marks) (d) Suppose 10 counters are open and one of them is randomly selected. Find the probability that more than 7 of them are busy and the randomly selected counter is not busy in a certain
minute. (3 marks) (e) The immigration hall is Called congested if more than 90% of the open counters are busy in a
minute. Suppose 15 counters in the hall are Open. A senior ofﬁcer checks the counters in a
certain minute. It is given that more than 7 of the ﬁrst 10 checked counters are busy. Find the probability that the hall is congested.
(4 marks) ZOOSAS—M SC 3—7 20 11. A manager of a maintenance centre launches an appraisal system to assess the. performance of technicians in terms of the time spent to complete a task. A technician can get 2 points if he takes
less than 2 hours to complete atask, 1 point if he takes between 2 and 4.6 hours, and 0 point if
he takes longer than 4.6 hours. Assume the time for a technician to complete a task is normally distributed with a mean of 3 hours and a standard deviation of 0.8 hour, and the number of tasks assigned to a technician follows a
Poisson distribution with a mean of 1.8 tasks per day. (a) Find the probability that a technician is assigned not more than 4 tasks on a certain day.
(3 marks) (b) Let p, be the probability of a technician getting it point(s) upon completing a task, where
1': 0,1,2 .
Find the values of pa, p1 and p2 .
(3 marks) (c) Find the probability that a technician gets exactly 4 points on a certain day. under each of the
following conditions: (i) 3 tasks are assigned, (ii) 4 tasks are assigned.
(5. marks) ((1) It is given that a technician is assigned fewer than 5 tasks on a certain day. Find the
probability that the technician gets exactly 4 points. (4 marks) ZOOBAS—M & 8—3 21 12. Ofﬁcials of the Food Safety Centre of a city inspect the imported “Choy Sum” by selecting 40
samples of “Choy Sum” from each lorry and testing for an unregistered insecticide. A lorry of
“Choy Sum” is classiﬁed as risky if more than 2 samples show positive results in the test. Farm A supplies “Choy Sum” to the city. Past data indicated that 1% of the Farm A “Choy Sum”
showed positive results in the test. On a certain day, “Choy Sum” supplied by Farm A is
transported by a number of lorries to the city. (3) Find the probability that a lorry of “Choy Sum” is risky.
(3 marks) (b) Find the probability that the 5th lorry is the ﬁrst lorry transporting risky “Choy Sum”.
(2 marks) (c) If k lorries of “Choy Sum” are inspected... fmd the least value of k such that the probability of ﬁnding at least one lorry of risky “Choy Sum” is greater than 0.05 .
(3 marks) (d) Farm B also supplies “Choy Sum” to the city. It is known that 1.5% of the Farm B “Choy
Sum” showed positive results in the test. On a certain day, “Choy Sum” supplied by Farm A
and Farm B is transported by 3 and 12 lorries respectively to the city. (i) Find the probability that a lorry of “Choy Sum” supplied by Farm B is risky.
(ii) Find the probability that exactly 2 of these 20 lorries of “Choy Sum” are risky. (iii) It is given that exactly 2 of these 20 lorries of “Choy Sum” are risky. Find the probability that these 2 lorries transport “Choy Sum” from Farm B .
(7 marks) END OF PAPER ZOOSASM & 3—9 22 Solution Remarks For binomial eXpansion [iii—ll
1. (a) (i) 1 “1+[i](ax)+2——2———(ax)2+m 414.63: 2 2! on (1+mc)2
2
___1_ﬁx+3a x2+ For any two terms correct.
2 3 ppl for omitting ‘...’
1&2.
22 2 Forboth
3a
__=b
3
2'7
a=—3 and b=—8— For both correct
(ii) The expansion is valid for Ix] <1 31
2 .'.———lhm—=1+1[—l—]+EZ +...
[1] 2 30_ 8 30
1—3 —
30
ENE F RHS t105375
9 300 or ,acccp .
J— 2529
1 a: 800 Accept3.l6125 and 3.161
3 _
2 (a) )2 —uy——1
dy dy
3 2— u—+ :0 M
y (it: [ dz; y]
d
._y= 2y 1;;
d“ 3y mu
1M
1A
2
(b) u=2x
111a=x21112 1M
i§=2xln2 1M
2
%=2x 2xln2 1A
Alternative Solution
u :21? = 63:21:12 1M
d 2
33: ex “12 2x1n2 1M
—2“‘2 2 111
— x 2 1A 26 Solution Marks Remarks
d d du
(C) _}.’.=._y..._
dx du dx
= 23’ 2x22x1n2 1M
3y —u
2 2
2x I+1
=—xy2 1A OR 2 W122
3y2—2x 3y2n2x
_ dz r+2 t+S x: [[53[;—;] 1.26”0‘1t]dr
5+2 t+5 =5.3m:+2)—1n(r+5)]—126“0”+C (since :20)
When 5:0 , x=0 .
.'.0=5.3(h12—1ﬂ5)—12+C
C=5.3(1n5—]n2)+12
616.3563
Le. x = 5.3 [1110‘ + 2) — 1n(t + 5)] — 126—01f +163563 (a) 3 = 5.3[;— — 1—] 1.26—0'1: OR 5.3[1n‘r+2[—ln —12e_0'1! +C OR x=+5.3ln (b) lim {5.3 [1110‘ + 2) — ln(r + 5)] — 1250*“ +16.3563} I—abee
=5.31im hag—121m 6‘0“ +16.3563
5—566 t+5 1—566 = 16.3563
i.e. the concentration of the drug after a long time = 16.3563 mg/L (a) P(A m B) = PM I B) * P(B)
k _______ 1A
6
(b) P(AUB)=P(A)+P(B)—P(AMB)
.‘.i=l+k—£ 1M
20 5 6
. 3
Le. k=——— 1A
10 (c) P(A' m B) = P(B) — P(A m B) =[%)%[%) 1M Alternative Solution
P(A’ m B) = P(A u B) — P(A) 1M =_, 1A 5. Solution (cl) P(A'rwB') =1—P(A U3)
9
_._1____ — 20 Alternative Soltion l
P(A' n B') = P(A’) — P(A’ n B) {la0) Alternative Solution 2
P(A’ m B') = P(A') + P(B') ~—+ P(A' u B') = 1294') + P(B') — P((A r3 B)') {or Alternative Solution
P(A') + P(B') = [I — +[
3 2 ¢P(A'UB’) since P(A'UB’)£1 Hence A’ and B' are not mutually exclusive. (3) Let $X be the amount of money spent by a randomly selected customer.
P(X > 30000) = 0.242 P Z>W =0.242
6000 30000 — a LPO<ZS
6000
= 0.253 ‘_ 30000  ,u = 0.7
6000
Le. t: = 25800 (b) The required probability
_ P(16500 «z: X < 30000)
P(X «4 30000) 16500 — 25 300 30000 — 25300
P ———— < z < —————————
6000 6000
1— P(X __>._ 30000)
P(—— 1.55 < z < 0.7) 1—0242
__ 0.4394 + 0.253 _ 0.753
a 0.9201 23 1M
1 lM 1A
1A 1A 1A 1A Remarks For P(A'nB') :2 0
Follow through For P(A') + P(B') at P(A'u 3’)
Follow through For standardization For P(16500 < X <: 30000) For denominator {Infilln H1“ Solution
(3) Since the mean of the 17 student marks is 74, k=74 (b) The required probability
_ c: er
(:5 3
105
z E (c) (i) The Standard deviation of the 18 student marks is 9.32737905 .
The corresponding interval is (74 — 2 x 9.3273 7905, 74 + 2 x 9.3273 7905) a (5534524139, 9265475311)
Thus, 55 is an outlier. ——'II'IIrIIIlII—HIIII'IIIIIIIFF——I'IIIIHI——Ii——iI—H The standard deviation decreases Easthe_e_atrerne_datum_§t_he_ ontiierli‘siremoﬁjed; 2
3x—2 3——
(a) lirn = lim 5" =3
x—>ee x + 2 x—>ee 1 2
+ _
x the equation of the horizontal asymptote to C is y = 3 .
.* lim 3x72 =+eo and lim iii—2 2—00
x—}—2_ x + 2 I—5—2+ x + 2 the equation of the vertical asymptote to C is x = 2 . (b) .
1 r
y = f(x) s y = 3
_...________________ I “3 __ _.______ ____ _ _ _.. _ _ _ 2 y = to)
x=—%
E l 3
E 3
43 —7 —a —5 4 «3 42 1 0 1 2 3 4 5 a
—1
—2 29 1M 1A 1A 1A
— 1A 1A .1A
1A
1A a Remarks For denominator OR 0.3860 Accept l d.p. or ab Follow through Awarded only when
correct For intercepts
For asymptotes Forshape (pp1) for origin an
of axes were all mi Solution Remarks . 3x—2
(C) (1) f(x) =
x+2
f ’(x) = W 1A For numerator
(x+2)2
__ 8
()t+2)2
So theequation of L1 is y——k=f’(h)(x—h)
y_3h—2= 8 2(x_h) 1M
h+2 (h+2)
(h+2)2y—(3h—2)(h+2)=3x—8h
8x — (h + 2)2y + 3/722 — 4k — 4 = 0 1 Follow through
(ii) (1) If L1 passes through the origin, then
8(0)—(h+2)2(0)+3h2 4h———4=0
h = 2 or :3?— (rejected since P lies in the first quadrant) 1A (pp1) if :5— was not rejected k = 3(2)_2 :1
(2)+2
2
Slope of L2 =~[—(—2—)—:8E]—=—2 1M
Hence the equation of L2 is y—l = —2(x—2)
Le2x+y—5=0 1A
. . 5
(2) The actIntercept of L2 13 5
So the required area
5
2 — ' —
= 23x Zm+l 22 (1) 1A+1A OR =~+j2(5—2x)dx
3. x+2 2 2 2
2 .
=J‘2 3.‘ 3 dx+—1— l (1) 1M For 3— 3
_ x+2 2 2 x+2
3
1
= 3x—31nxI2 2 +—
[ l “a 4
3
= 177+31ng— 1A OR 1.0063
(10) 30 Solution Remarks
3. (a) (i) N’(r) = A7“ (2‘ 2 0)
1+ he—
in —29— —1 = —k2‘ + In h
N'O‘)
(ii) lnh = 1.5
h=35
m 4.4817 (correct to 4 d.p.) Either One
_k=15—0
0 — 7.6
k = E
76
m 0.1974 (correct to 4 d.p.)
(b) (i) v = 4.5 + em”
£11 = 0.2302! 1A
(12‘
20
NU) = dr
J1+ 4.5e'0 2‘
= j 100 (0.2.?0 25:1:
30 2’ + 4 5
= £99.11, 1M
v
= 100 Inlvl + C
=1001n(4.5 + 302:) + C 4.5 + eo'zr > 0) 1A
Since N(0)=50 .so C=50—1001n5.5 1M
0.2!
i.e. N(t)=1001n—£—L—5+e—+50 1A
5.5
(ii) (1) M(20) = N(20)
0.2(20)
21 eon5145150290) + b =1001ni§i§__+ 50
0.2 5.5
rho141.2090 1A
(2) Consider M'(t) — N'(t)
= 210— 4.53'02' ) — #39? 1A For the 15't term
1+ 453— ' I
1— 425.25 "04"
1+45€ ‘
. . ~04: 1
..M(r)—N(r)>0 when 3 < 1M
425.25
i.e. I>Ei2§;25#15.1317 1A
0.4
Since M(20) = N(20) and M(t) — N(z‘) is increasing when t > 20 ,
so M(t) > N(t) for I)» 20 .
Hence the biologist’s claim is correct. 1A Follow through 31 Solution Remarks 10
9. (a) (i) L) 2136— 1+t2dt
s%.2%[4/1+02 +2\/1+2.52 4201+52 +2\i1+7.52 43114—103] 1M
s1.305132044
61.3052 1A So the increase of temperature is about 1.3052 “C . (11) it‘s—OW] = 44—: d’ 40 1+ :2
2
L[~—i—\il+r2]=—1 1A
d32 40 3.
40(14143)2
> 0
Hence it is an over—estimate. 1A Follow through
(b) (i) 100(111 x0)2 —630111x0 +1960 = 963
500630?43151115:0 +496=0 1A
31 16
lnxo =— or —
10 5
x0 6 22.1930 01‘ 24.5325 1A
(ii) w,(x): 2001nx_63:0_ 1A
2: :1: Wm < 0 when 200 lnx  630 < 0 (2* x 2 22) 1112:: <: 3.15
1.6. 22 g: x < 6315 s 233361 1A Accept x < 23.3361 dW dW dx
(111) —— = ———+
dz dx dr
2001 — 2
: us: 630 1+: 1M
x 40
When {=0 , x=22.
_d_W_ _2001n22630_3/1+0 1M
dz‘ {:0 22 40
m —0.0134 1A
Hence the rate of change of electricity consumption at 2‘ = 0 is —0.0134
units per year.
(iv) The electricity consumption at 2‘ = 10 is approximately
W(22 +1.305132044) 1M
= 1000112331051820441)2 — 63011123305132044+1960
#967,7502 units 1A 1 =10 is x < 23305182044.
Moreover, W(x) is decreasing for 22 3x < 23.3361 by (b)(ii). Therefore the actual electricity consumption is larger than this estimate. 1A Follow through . Since the estimate in (a)(i) is an overestimate. the actual temperature when }
1M 32 (b) (d) (e) Solution Remarks The required probability
__1_ 3 903—3 9 + 3 919—39 3 922—3 9 3 93 2'3 9 1M for cases correct
_ ()1 11 + 21 + 31 1M for Poisson prob:
m 0546753239
43 05463
The required probability
2 1— P(no busy counters are found after the 4th counter is checked)
a 1 — (1 — 0546753239? 1M
a 0.9573 1A
Alternative Solution 1
The required probability
a C14 (054675323 9)(1 — 0.546753239)3 + cg (0.546753239? (1 — 0546753239? + C34(0546753239)3 (1 — 0.546753239) + (0546753239? 1M 1M for Binomial proba'
m 0.957 3 1A Alternative Solution 2
The required probability m (0.546753239)+(1— 0.546753239)(0.546753239) + (1 — 0546753239? (0.546753239) + (1 — 0546753239? (0.546753239)
a 09573 1M for Geometric prob The required probability
a (054675323 9)10 + (:30 (054675323 9? (1 — 0.54675323 9)
10 3 2 1M for cases correct + C3 (0546753239) (1 — 0546753239) 1M+1M M for Binomial pm].
a 0.096004444
9:: 0.0960 1A
The required probability
a €310(0.5410753239)8 (1 — 0.54675323 9? x % 1 + C910(0.546753239)9 (1 — 0546753239) x E)— + 0 1M+1A 1M for form correct
m 0.0167 1A
The required probability (0546753239)10 ><'[(0.5467S3239)5 + Cf (0546753239? (1  0.546753239)] + ago (0546753239? (1 — 0.54675323 9) x (0.54675323 9?
0096004444 1M for denominator 11: 1M for numerator fom
1A for numerator corre W Alternative Solution The required probability 1M f r demo . t
o mma or us. . 1M for numerator forrr.
1A for numerator corn: ﬂ (054675323 9)15 + (311.? (0.54675323 9?4 (1 — 0546753239)
0.096004444 m 0.0163 Solution Marks Remarks 11. (a) The required probability
1M for P(X 5 4) T + "‘17—— + “T + + IMHM 1M for Poisson probability
N 0 963'593339 ' ' ' ' with correct ,1
a: 0.9636 1A
4.6 — 3  . .
(b) p0 = P(Z > 0 3 ] = P(Z :> 2) = 0.5  0.4772 = 0.0228 1M For standardlzatlon
p2 = P(Z < 20—33] = P(Z <: —1.25) = 0.5 — 0.3944 = 0.1056 1A For any one correct
p1 =1— p0 — p2 =1 0.0223 0.1056 = 0.3716 1A For all correct (c) (i) The required probability = C13 p2 p12 + C13 pzzpo 1M 1M for form correct
= 3(0.1056)(0.3710)2 + 3(0.1056)2(0.0228) e 0.241431455 *4 0.2414 1A (ii) The required probability 1
= Cgpzzpoz +ﬁp2p12p0 +1714 1M 1M for form correct
= 6(0.1056)2(0.0228)2 +12(0.1056)(0.3716)2(0.0228)+(0.8716)4 1A
a 0599107436
a 0.5991 1A
((1) The required probability
2 —1.8 3 l.3 4 —1.3
L(0.1056)2 + 1'—33—(0.2::11431«455) + 1'8;(0.599107436) 1M for any 2 cases
:3 2! 3! 4! IMHMHA 1M for denominator using 1
0963593339 1A for all correct
a 0.0333 1A 34 Solution Marks Remarks 12 (a) The Tech1399 Pmbability 1M for cases correct :4 0.0074973 63 (Can be awarded in (1
# 0.0075 1A (b) The required probability ,4. (1—000749736204(0007497363) 1M
:3 0.0073 1A Cf (0.0074973 63)(1 — 0.0074973 63)k_1 + Cﬁ‘ (0.007497363)2 (1 — 0.0074973 63)" 2
+  u + Cf 1(0.0074197363)k'1(1— 0.007497363) + (0.007497363)k 5 0.05 34
("2
3...»! Alternative Solution
0007497363 +(1— 0.007497363)(0.007497363) + (1 — 0.0074197363)2 (0.0074973  3) + + (1 — 0.007497363)"‘1 (0007497363) 5» 0.05 1— (1 — 0007497363)" 5» 0.05 0992502636" < 0.95 k 11] 0992502636 < in 0.95 k > Jig—'31— m 6315332223 1110992502636
Hence the least value of k is 7 . (d) (i) The required probability
=1—[(1 ——0.015_)““0 +C140 (1—0015)39(0015)+C§0(1—0.015)33(0015)2] m 0.022069897
m 0.0221 (ii) The required probability
= [C3 (1 — 0.007497363)‘a (0.007497363)“ 10:12 (1 — 0022069397)” (0.022069897)2] +[C13 (1 — 0007497363)"(0007497363)][C§2(1—0022069397)”(00220693901 +[C§ (10.007497363)5(0.007497363)2][C32(1—0022069897)”(0.022069397)” m 0.037154730
m 0.0372 1M for any 1 case co:
1M for all cases corr (iii) The required probability
[C30 — 00074973633(0.007497363)0][C£2(1— 002206939010(00220693902]
" 0037154730 1M for form correct
1M for denominator 1 .43 0.6517 35 ...
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 Spring '11
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