Download P(A

ECE 109 Midterm
SOLUTION
Name: _
_ _ _ _ _ _ __
May 8, 2013
PID: - - - - - - - - - - - - - - - -
PROBLEM 1- ECE109 Midterm
1. Let A and B be events that satisfy
P(A)=0.4
P(A U B))= 0.8.
P(AnB))=0.3
Evaluate:
( i) P(B)
( ii) P(Ac)
(iii) P(AnBc)
(iv)P(Acn Be)
An answer not supported by appropriate reasoning will not receive credit.
P(AUB)
(L')
11
o.e
~- t
)
P(A) + PlB)- P(AnB)
=.
u
+-P(S)- o.3
= o-4-
P( A') = l - Pl f\
-:;
f(A') ::
PC A-B)
==
=
0 .7
( ii) P(Ac) =
0 .6
o. 7
o. b
P(AJ- PtAnB)
fl
"
( i) P(B) =
P( B)=
\\
=.. 0.4- o-3
p(A()f3') -
,,
=7'
)
''
P(A')
::: I - o.II t
p(AnB')
tl
II
- I- P( AUB)
l\
(iii)P(AnBc)=
D. I
=7P{Anec)=o.l
SOLUTION
Name: _ _ _ _ _ _ _ _ __
PID: _ _ _ _ _ _ _ _ _ __
PROBLEM 2- ECE109 Midterm
2. A random variable takes on the values {-1, -2, 1, 2} with the probabilities
P(X=-2)=
1
6
,
P(X=-1)=
,
3
1
P(X=1)=
,
3
1
P(X=2)=
1
6
.
2
Evaluate E[X] and E[1/X].
An answer not supported by appropriate reasoning will not receive credit.
k
'X Itt
(
-2.
2
3
-I
I
2
t
-x~
4
--
I
-I
113
I
I
'/ 3
+
-2I
y~,.
I
2.
Pflt-= PCX~Xp)
......
116
'/6
E[Xj :
EL'/tj ~
2
I E[X ]
~
1.
E[l/X] =
0
SOLUTION
Name: _
_ _ _ _ _ _ _ _ __
PID: - - - - - - - - - - - - - - - - - - -
PROBLEM 3- ECE109 Midterm
3. In a course on probability 80% ofthe students passed. Ofthe students who passed, only 1%
did no homework, while 12% of the students who failed did no homework. What is the
probability that a student who did no homework passed the course?
An answer not supported by appropriate reasoning will not receive credit.
PC fGLd) = o.:z_
PC pa.~c;) = o. e
P(v.ob-1w/f-o.,l )=
P( Vl~ t{W / pa~') = o.o1
:. PC po~~)~
~o Hw)
o,l2..
P(t~o HW}
~ p(po~s)
p(
¥\(.)
l-tW)
p(v." Hw} pa~7)
Tot&( I Pro b_ -7 P( Jlo Hw) _ P( vw Hw) p()~'?). p[ po~?)
-r PC VlO HwI fo.,J). f (~()(/)
TN2of'()m
0 . 0 ,)( o.e + o.tzx. o. 2
-_,
IP(passl no
HW)
~
() • ')..
!7
0, 03'2..
Name:
SOLUTION
PROBLEM 4- ECE109 Midterm
4. A game is played as follows: A fair (unbiased) coin is tossed and the player must guess
whether each toss is a head or a tail The game continues as long as the player guesses
correctly; it ends at the first wrong guess. If the game has lasted longer than n plays (tosses),
what is the probability that it will last longer than n + m plays where m 0!: 0.
rJfJ
P(NP >r1) = LP(NP:p)
Jrt.:,..+l
~
{r±)k
~=
,.+,
-
-
:: &)n
P(last longer than n + m playsjlasted longer than n plays)=
(k) ~
ECE 109 Final Exam
SOLUTION
Name: - - - - - - - -
June 12, 2013
PID: _ _ _ _ _ _ __
PROBLEM 1- ECE109 Final Exam
I
I
1. Two events A and B satisfy P (Be A c)= 1. Evaluate P(A B).
An answer not supported by appropriate reasoning will not receive credit.
p ( B' IN)= P(B enA")
;n-1. =) P(B"nA')
P[A')
PC scnAc.)
" P{A')
= 1- P(AU8)
= 1- PCA)-PCB)+P(Ane)
p( ,4 c)
1- P[A)-
=- 1 -
p((i )
PCB)+ P(An9) = t- PCA)
{V
p[A n B) ~ PC 8)
.-
IP(AIB)~ 1
P(A I B)
=-
P{Anr3l = 1.
PC f3 J
Name:
SOLUTION
---------------------
PID: ----------------------
PROBLEM 2- ECE109 Final Exam
2. Let X be a Poisson random variable
CoS nn-
=
1 (
2
;..,n e-A, n = 0,1, ...
P(X=n)= {
n!
0
~
'
Determine the mean and variance of Y = cos(1tX).
An answer not supported by appr:/riate reasoning will not receive credit.
£:: ( C ~S ~>11 J
"-
5_
V\::.0
CC$>11f _} ~-
rll
:: {;-0
V\o ..
e}
f01 -~
n!
- 2,A
e =e
_ t"rt
+€
n)
~ c~~t +(-·)~
=
n<O.
€
t~rr
f-1 )n
Name:
SOLUTION
------------------------
PID: __________________________
PROBLEM 3 - ECE109 Final Exam
3. The random variables X andY have the joint probability
2 n-2
_
_
_
{
P(X- n, Y- m) -
p q
O,
, 1 s m s n -l,n ~ 2
h
.
ot erwzse
Where p + q = 1 and 0 < p < 1. Evaluate the marginal probabilities of X and Y.
You must specify the allowable range of values X andY can have.
An answer not supported by appropriate reasoning will not receive credit.
~
P(A ~ ~ )
'L FtX:= V1) Y= W)) -
"=-
~~
~ p2~~-2.
u
L
~::-ob
P('t:~)
(
=
)
P(X=n)= \.Vl-1
Range ofn =
2..
>
pl.t£IT tt-l.
P1J' ""-1
P(Y=m)=
Range ofm = ~
1
Name:
SOLUTION
-----------------
PID: _ _ _ _ _ _ _ _ __
PROBLEM 4- ECE109 Final Exam
4. The random variables X and Y have the joint density
f
X,Y
(x
,y
)=
{
2,
0,
0 :s y :s x :s l
h
.
ot erwzse.
Evaluate the covariance of X and Y.
I
Cov[X,Y]
~ ~~
SOLUTION
Name: _ _ _ _ _ _ _ _ __
PID: _ _ _ _ _ _ _ _ __
PROBLEM 5- ECE109 Final Exam
5. The random variables X andY are independent with densities
fx(x)
=
A.e-A.x 'x c::: 0
o,
x<O
{
f
_ {
y(y)-
o,
yc:::O
A.y'
0
A.e , y< .
Evaluate the probability density of Z =X+ Y. You must indicate the allowable values of Z
for this density.
An answer not supported by appropriate reasoning will not receive credit.
- 2:+x(~-;kcr>otr
V"tU
?t- ho. v..Q
~-1¥ ~0 0..-A)
u
-g
t-N"'
.[:2[})
l
==
T~VI
~
;~I
r
!It
not ,s-
lf-})f£("'f)dq :
r <o
h r1 ) =
c:>
tfr rt -r»t i )rJ.3
l
11- ~0
u
o.d y'-o
Name:
SOLUTION
--------------------
PID: _ _ _ _ _ _ _ _ _ ___
PROBLEM 6- ECE109 Final Exam
6. The random variable X has the probability density
A.?xe-AX, X~ 0
f (x)- {
x 0
x<O.
'
Evaluate the mean and variance of X. [HINT: You may find the characteristic function
helpful.]
An answer not supported by appropriate reasoning will not receive credit.
~x (1L) "
E[
e'"IAX J = ~.ol~x { lt( :~) t~} r)-.J.
(. ~ )2.. ~I
~}- .i u
IE[X]
~
Var[X]
=
2./>.
2/"' :L-
~---
T-Atrcd U: o Ill
U~v l F" rt~ttcJl as Po.~
t.t,
w~Th
~.::.4--
Name:
SOLUTION
-------------------
PID: - - - - - - - - - - - - - - - - - PROBLEM 7- ECE109 Final Exam
7. The random variables X and A have the joint density
>1
>0
- ae -ax ,x_,a_
f x 'A (x,a) - { 0 , otherwise
For a fixed value of A= <lo• evaluate the probability P(X > Xo IA = <lo), with Xo > 1 and
<lo > 0.
An answer not supported by appropriate reasoning will not receive credit.
DO
P( g: >;t a I Ae (j. rJ) ;;;
~ f.X: IA.~~ IA=do) ol.,x
~(}
{~~A (X; o<o~ _
f,Pr ((/(rJ )
(:)()
F.~~ 11y pex >"1 I.A::,(") ::
0
l o<a ed-o()l.cl~
'J.o
- o<o("Xo-1)
=
e,
'Xo > 1_
Name:
SOLUTION
---------------------
PID: - - - - - - - - - - - - - - - -
PROBLEM 8- ECE109 Final Exam
8. The random variables Xk, k = 1,2, ... are independent with probabilities
P(Xk~n)~ (A.,)"
e-At,n-0,1,...
n!
.[!XII (1<)
Ak = (1 I 2) ·
""
"Joe 0.¥ e-~~
ro
k
and
..
,r..u~
()()
::
.e_!P-( .e i-J -
1)
Evaluate the characteristic function ofXk. Now consider the sum
Evaluate the probabilities P(Z = n). [HINT: You may find the characteristic function useful
in determining P(Z = n).]
~
~(i-)
Jo_::;/
-
_,
i;--:a.--fv--om
u c;~fo [
f" v- ~ vla~'
l
P(Z = n)
=
~
e}
-l
1-\:: O, IJ
• ·'
ECE 109 Midterm
May 7, 2012
Solution
PID:
PROBLEM 1-ECE109 Midterm
1. A communication channel transmits binary signals. The input to this channel is either a zero
or one with the probabilities
P(ln=1)=315
P(ln=0)=215.
During transmission errors may occur. The probability that the output is a "0" when a "1"
has been sent is P( Out = 0 jln = 1) = 1I 3. The probability that the output is a "1" when a "0"
has been sent is P(Out = 1jln = 0) = 1/3.
Evaluate the following:
( i) The probability that the output is a "1."
(ii) The probability that the input is a "1" given that the output is observed to be a "1."
Ananswer not supported by appropriate reasoning will
~ot
receive credit.
:: P( 0 ut- =' }t~ =CJ) P( I ~ o )
1
:::
+ p(Out= tl Lll'o c: t) P(I,,., = I)
-= ( ~) . ( -?) +- ( l - ~) (
--
( i) P( Out= I)=
8/ I J
(ii) P(Jn = ljOut = 1) =
3/+
} )
Solution
Name:~~~~-------
PID:
PROBLEM 2- ECE109 Midterm
2. The spaceS= {1, 2, 3, 4, 5, 6, 7, 8} is an equiprobable space. That is, each elementary
event has probability 1/8. Consider the three events:
C={1,3,5,7}
B={1,3,4,6};
A={1,4,7,8};
Determine if:
( i) A and B are independent
( ii) A and C are independent
(iii) A, B, and Care independent.
An answer not supported by appropriate reasoning will not receive credit.
I
I
PCA)= 2:
AnB =f1)4j
An C =~I)
P(AtlB)::
7i
AnB()C : ~
11
\
PC)= 2
1,
fJ(Anc)=t
P(AnBtlC)= ~
p[A fl B) = Pl A) P( S)
P(AAc) =
(
P(S)= 2
PCA)P[c)
A avrrl g
A
a'Ad
c
a.~
O.V'(>
l
"'ckf-R w~t
-.~dife"'htAt
P(An a!) C)= PCA)Pl~)P[c) A) Bo...,..J c av-p L~J.ep.eVIkll{t-
( i) A and B independent?
( ii) A and C independent?
(iii) A, B, and C independent?
No
No
No
Solution
PID:
PROBLEM 3- ECE109 Midterm
3. A random variable X takes the values {-1, 0, 1, 2} with the probabilities
P(X=-1)=
5
12
,
P(X=0)=
1
4
,
P(X=l)=
1
4
,
P(X=2)=
1
12
.
Evaluate (i) E[X] and (ii) E[X\
An answer not supported by appropriate reasoning will not receive credit.
~,-,1).
P(&:. ::.
~):::.
)
'J4 J
{ '/4 J
jQ,_::.- {
kl.-:o
la.~L
1/tl}
2.
S(XJ _ 2_ ~
k~-l
P( X:=-~)
kz.:::- :z...
= (-t )( fi) t- (o)(~)t{t )(~)~ )(T2)
:.0
~
2
~ ~"l. P{X:~ h) ~ (-1) {, 2 )
la.:.-1
;:: 1
( i)
I
E[X] =
2
(ii) E[X ] =
0
1
":l..L
')
":1..1_)
1....'
+-(o)~(;;
+( 1)~t(2)(
12.)
2
Solution
Name:~~~~~--------------
PID:~-----------------------PROBLEM 4- ECE109 Midterm
4. A sequence of independent trials (success or failure) continues until a success occurs. The
probability of success on a trial is 113. What is the probability that three or more trials are
required before there is a success? That is, if X is the number of trials until a success occurs,
find P(X;;?. 3).
r
An answer not supported by appropriate reasoning will not receive credit.
p :.
'0-.,
P(X = Vl)
V'O
~- 0
1-
p
f
S tiCGQ ~ ~
=· pro b. a f
f ~ ...- I )
"'=
f-a.. I"~
L J2., -·.
V1-l foulvt'€S ~~
Q VI
P(X;;?. 3) =
Ill
t"lfJ
rrL td
SOLUTION
ECE 109 Final Exam.
June 13, 2012
Name: ________________________
PID: ------------------------
PROBLEM 1- ECE109 Final Exam
1. Consider the random variable X which is uniformly distributed on [-A, A]
fx(x) = {
1
--, -AsxsA
2A
.
otherwzse
0
'
Evaluate the probability that
lXI is greater than 1/10 of its maximum value P(IXI >A I 10).
An answer not supported by appropriate reasoning will not receive credit.
p(I X\ >Ajto)
==
5{.z: ( J..r -- dA ) d.;><
y.)
I-xl >Aflo
»
A/to <I'X) ~A
A
J_
It
)c4
Aj1o
-
t(A-
~)=
'1/to
SOLUTION
PID: _ _ _ _ _ _ _ _ _ _ __
PROBLEM 2- ECE109 Final Exam
2. The time to failure of a computer's hard drive is denoted by T. Let P(T > t) =
e -A.t.
What is the probability that T exceeds its mean value P(T > E[T])?
An answer not supported by appropriate reasoning will not receive credit.
Fr(t)::
1- P( T >t)
fT I-t) = J fr{-t) = ~
==
I-
/)-r.t-t:--
7 i- ~ o
>. .e-H) t <:o
0;
tL.O
00
- )A+i>-:Jto
~-
{)()
P( T > ELTJ) = p(T>-})-
_,
- -e
I
P(T > E[T])
~
--1
e
- t.t-
) H? olt
Y;..
SOLUTION
Name: ---------------------PID: ------------------------
PROBLEM 3- ECE109 Final Exam
3. The random variables X andY have the joint probabilities
2 n-2
P(X = n, Y = m) = {
P q
0,
'
1s m <n
.
otherwlse
Evaluate
( i) P(Y = m)
( ii) E[Y]
p
( i) P(Y = m) =
( ii) E[Y] =
ljf
{
Pcu-WI-J)
0
)
W\ ~I
oi'-~11' vvt~
O<p<l and p+q=l.
SOLUTION
PID: _ _ _ _ _ _ _ _ _ _ __
PROBLEM 4- ECE109 Final Exam
4. The cumulative joint distribution of the random variables X andY is given by
F
(
X,Y
x,y~o
)- { 1-e-ax-e-aY+e-a(x+y)
x, Y o,
'
otherwise
with a> 0. Are X andY independent?
An answer not supported by appropriate reasoning will not receive credit.
P
i:=
'i(<t) -:: '~J'~(<IJ;"j)
fx- (I') f<;?:l"'f)
"t
=
,_ear )
{
=
C)
-!lll<) (I --ioY)JJ
e-ar
-t--€
-a~
{
}
I --€
0
-C<('tf;)
)
)
C)
t
(A.V\,)
I
X and Y independent?
A
fA.VIJ
F.rC~<J F~ L}')
:l ().'(\(! t'Vt.ckp.e,vtthHt
=.
No
f
hP V'v.ltS-{>
C l..ea r-ly
fx> T C~<' ~)
:tJ
~0
SOLUTION
PID:
PROBLEM 5- ECE109 Final Exam
5. The random variables X andY are independent with densities
fx(x) = { ;..,e-AX' X 2: 0
0,
x<O
f ( ) = { f-lC -f-ly' y 2: 0
YY
0,
y<O.
Evaluate P(X > Y).
An answer not supported by appropriate reasoning will not receive credi
PC X > )'")
IP(X>Y)~
1-
=
~
"-
SOLUTION
PID:
~-----~-~·
-~--·~~
---
PROBLEM 6- ECE109 Final Exam
6. The random variables X and Yare independent and uniformly distributed on [0, 1]
fx(x) = { 1, 0 s x s 1
0, otherwise
Evaluate the density ofZ =X- Y.
fy(y) = { 1, 0 s y s 1
0, otherwise.
-l~~L./
c/Dcvrlv
-c.
I
An answer not supported by appropriate reasoning will not receive credit.
V"VSf
sa:h~ f.y
(J
~l
St wr,ltu~.er;~~
f= I
.
().'otJ ()f1
-
+-r6'
f:
~~-
t-1]1 )-l!:i61
0)
oThav-'wt ¥
SOLUTION
PID:
PROBLEM 7- ECE109 Final Exam
7. The random variables A and B are independent with densities
fB(f3) = { 3 I {34, f3 t:d
0
otherwise.
fA( a) = { 1/3, 0 s as 3
0,
otherwise
'
Consider the new random variables X= A+ BandY= A- B. Evaluate the covariance of X
andY.
An answer not supported by appropriate reasoning will not receive credit.
Cov (K")
~J::
E[XYJ - ,;:Lg-J EL~J
= (Are)(A-8)] -(~lAJ+ELBJ)(£Lil]-ELt3])
El
- E L A:1 - t L8 :2J - ( El PU)
2
+{EL8J)
2.
_ V(). rLAJ - VetrLf?J
s [Aj =
3
H-
D<
do<=
i
()
3
E:L A'J=
~ Vu-r[A] - El Pl J -(ELAJ)
2
-z
J?x'do<-
3
="2f
00
()(()
~ [ 13] = ) 3 f~ =-%::I
V o.rlBJ
I
Cov[X,
Y]~
(34-
ElB?J = f 3{3~ :::3
= E LtrzJ- ( £L BJ)~ =-
Q
=3
I
t
(3
Name:
SOLUTION
---------------------
PID: _____________________
PROBLEM 8- ECE109 Final Exam
8. Let X be a Binomial random variable
_ O, 1, ... , n,
_ m ) -_ ( m
n ) p m( 1-p)n-m , mP (X-
O<p<1.
( i) Obtain a closed-form expression for the characteristic function <l>x(u).
( ii) Evaluate the mean and variance ofX.
An answer not supported by appropriate reasoning will not receive credit.
h
-
~g:(-z-t):::. £l.fl~uXJ = 2_ --e"u~P(X== WI)
=
{
(
W\:::
B
1VI
OWltC:d
~)
WI
(
::.Q
p~;~)~Ct-P)V\-W7
0
( '- p +-- r-e'/Yl
Svm
f-V'o t1, , ' I) ~i.e ~ v~l
11
fo r»tJIO'~
-
Vl
t:- IK'J =-
p
(f:Ji-,~l"Jtl/wo =ft/( i'..p{) (l-pt- r/) .,_,)
r
7A,::; iJ
= V1p
+VI(Vt-l)p
"\j_·f( ~
2..
(ii) E[X]=
(1- P -t-p-e~u)VJ
Vl
f
Var[X]=
>'
{•-•)fe''")(i-p~p e~)
u-=.o
VaV'[KJ ~ £:L%~-(f:liJ)2_
(i) <I>x(u)=
2
V),
P( [- p)