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# answersbookchapter6 - Chapter 6 P1(c)P{X=1 Y=1}= P(l 1 3i...

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Unformatted text preview: Chapter 6 P1(c)P{X=1, Y=1}= P{(l, 1)}: 3i, P{X=2, Y: 5}=P{(2, 5) (,5 2)}: 32—6— — 1—18, etc so in general —— iflgi=j56 P{X=1',Y=j}= 8iflgi<j£6 0 otherwise. “I P2(a)P{X1-1X2=1}=P{X1=1}P{X2=1|X1=1}___ﬁ=%16’ P{X1=1,X2:0}=%. %_1:_%1 P{X1=0X2=0}_ 34,}. %=%%, oo 00 P10 (3.) P{X < Y} = f fzq e‘(z+y)dmdy = f U e_(3+y)dm) dy = f (e—v — e—ZV) dy = o o o [—e'y + #49]: = -;- (Or one can get this answer immediately noticing that in this example f(:z:, y) is symmetric with respect to a: and y, and thus P{X < Y} and P{ Y < X} are equal to each other and together give 1.) P12 Assume that a person entering the store is a man with probability p. It is shown in Example 2b that the number of men X and the number of women Y entering the store in one hour are Poisson variables with parameters A13 and )1(1 — p) Therefore P{X s 311/ = 10}: P{X = 3}: 2 P{X = i} = z: e-APLEL= (e). i=0 i=0 In particular, with A = 10 and assumingp = 0.5, we get (*) = e— 5(1+5+%+1—§-5-) 'e‘: 0.265 P13 Let X be uniformly distributed on (15, 45) and Y on (0,60) (measure time in minutes after Noon). Then P{the ﬁrst to arrive waitss 5min} = P{IX — Y| 5 5} = P{X — 5 5 Y S X+5} =(area of KLMN)/(area (-1 ABCD) = %. P{the man arrives ﬁrst} = P{X < Y} = % by symmetry. P17 1/3 (Under most reasonable assumptions on what “three points are selected at random (and independently) on a. line...” might mean.) P22 (:1) No because f (any) = a: + 1; can’t be represented as a product of a function of a: and a function of y. (b) When an is inside (0,1), kav) = T fx.v(\$.y)dy = Ohm + Indy = x + %- When a: is outside (0, 1), fx(:n) = 0. 1-y (c) P{X+Y < 1} = Hm.“ f(x,y)d=cdy = {1 ( J (m + ind-x) dy = of l—aﬁdy = l 3. P26 (3.) FA,B.c(a,b,c) = P{A 5 a,B 5 (1,0 3 c} = P{A 5 a} -P{B E b} . P{C S c} = abc (when 0 _<_ a,b,c S 1); fA.3‘c(a,b,c) = 3% E 1 (inside the I l (b) P{roots are real} = Pm? 2 4A0} = fff 0 < a b c < 1 1 -dadbdc = cube). (22 2 4m: ffusa’CSIU—Wlacﬁadc: C «inc 3 1 v... 1/4 1 1 ﬁ 4 . f (f(1 — v4ac)dc)da + f f(1 — \f4ac)dc)da = o 0 1/4 0 1/4 1 f( ~5~/5)da+r‘§ f ida= 0 1/4 Med” when a: 2 0 — 0 when a: < 0 FZ(Z) = P{Z S 2} = P{X1 S 2X2} = If 0 < 31,372 < 00 fx1(\$1)fxz(\$2)d31d\$2 = :61 S 22:2 P2'i'l The density function of X,- (where i = 1, 2) is f X.- (as) - 00 22:2 00 f f Ale—A'IlAge‘Mndxl dzrz = fAze”"”= (1— e"“”“) deg = —1——’\ if z. In a o o 2 1 particular, P{X1 < X2} = P{Z < 1} = F20) = Ail-31‘ 1This problem was different in the Seventh edition. Sorry, I did not notice it earlier! ﬂat 1 3 P29 Assume that the gross weekly sales X at different weeks are independent . variables (with 114—: 2200 and 17,-: 230). . (a) Let X: X; + X2. Then X is a. normal variable with 11:11.1 + 11.2- — 4400 , and cr— — 1/‘71 + or;— - 230ﬁ~ ~ 325. Let Z be a. standard normal variable. Then P{X>5000}= P{Z> W~L85}=1—Q.(185 :} —0.~9678 0.032 (b) Let p be the probability that sales during one . wk exceed 2000 Then p = P{X > 2000} = 1 — P{X < 2000}: 1 — P{Z 5 Egg ~ —0.87}= 1—Q(— 0.87) )=1—(1—-Q(0.)87) Q=(0.~87) 0.8078 P{weekly sales exceed 2000' in at least two of the three weeks}: p 32+31) (1— p)~ 0.903. P40(a)py(1)=P{Y=}= 1+1 GivenY=1 px|y(l|1)=1’1-§-= till“ ﬁlh‘ p1121= P{Y: 2}= i + is = 3-- 1", 1U“ 2.?X|Yl(2|1l= =%- 1 Given Y— = 2 pry(1|2)=3—'1PX|Y(2|2)=§ = E (b) X and Y are not independent because Pxfy is not the same given Y: different values y. (c) P{XY<3}=1, P{X+Y>2}=0 P{X/Y>1}= p(2 1)=-. P41 (a) fx(w)= f f(:0 y )dyzfxe IW+Udy=e “ 191/01): _f f( 3;, y)d:r= :fo\$€'z(y+l)d\$= W 2(11 “3 fX|Y(miy)= 7%??= #— = :1:(y +1)25 (”1) —-a(1r+1) fYIX (yi\$)= fxa: y) = me e—: (x; t/Z’ (b) IfZ = XY then Fz(t) = P{Z S t} = ffzyﬂ f(as,y)dmdy = f (f me‘m(y+1)dy) da: = — 0 0 — —:ce‘“ :fe-ru— e“)dx = 1 ~e-i so fz(t) = dag-(1)) = e-1 P44 P{largest of X1 , X2, X3 > the sum of the other two}: (for symmetry): 3 P{Xg > X1+X2}— 3fff 0 (3113;21:33 < 1 1d\$1dscgdx3~ — 3ff 0 <\$11\$2u < 1 (1- \$32\$1+\$2 ml+m2<1 1— :t] (.781 +32))d\$1d\$2— — 30f( f( 1 -— (121 + x3))d:1:2) dx1= 30f jig-L615“ = 3'3 =2. 0 Geometric interpretation: this probability = the sum of volumes of three tetrahe— drons inscribed into the unit cube. _ ,._, total of trials T3 Take L — 0/2" Then 7r N # of was wEen nﬁle meets a. line T6 fx+Y(t) = ﬁFx+Y(t) = %P{X+Y S t} = ad: (1? (irks/(\$1 Eddy) d3) = f Iii (tlexyws 10611!) do: = of fx,y(z,t — x)d:c ...
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answersbookchapter6 - Chapter 6 P1(c)P{X=1 Y=1}= P(l 1 3i...

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