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EECS 126 Midterm 1 Solutions: Probability Concepts, Exams of Probability and Statistics

Bharat Ratna Dr. B. R. Ambedkar UniversityProbability and Statistics

The solutions to midterm 1 of the eecs 126 course offered by professor tse at the university of california, berkeley, in the fall of 1998. The solutions cover various probability concepts, including independent and mutually exclusive events, conditional probability, and bayes' theorem.

Typology: Exams

2012/2013

Uploaded on 04/01/2013

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UNIVERSITY OF CALIFORNIA
College of Engineering
Department of Electrical Engineering and
Computer Sciences
Professor Tse Fall 1998
EECS 126 — MIDTERM #1 Solutions
1a. (i)
independent .
(ii)
mutually exclusive .
(iii)
.
( reads, “If event occurs, then event must occur.)
(iv)
.
b.
since .
2.
Let be the lifetime of the machine picked and event that machine is picked.
Note: not necessarily exponential, as some of you have assumed.
3a.
Note: Some of you wrote:
a NO-NO!
EF
,
PEF
()⇒
PE
()
0.4==
EF
,PEF()⇒
PE F
∩()
PF()
------------------------
0
==
P
EF()
PE F
∩()
PF()
------------------------
PF
()
PF()
------------
1
===
FE⊂
F
E
P
EF()
PE F
∩()
PF()
------------------------
PE
()
PF()
------------
0.4
PF()
--------
----
===
BA()
()
()
PA()
--------------------------------PB()>=
PAB
()
PA
()>
X
A
i
=
i
P
A1Xt≥()
PX t
≥
A
1
()
PA
1
()
PX t≥()
----------------------------------------------=
= PX t≥A1
()PA
1
()
PX t≥A1
()PA
1
()PX t≥A2
()PA
2
()+
----------------------------------------------------------------------------------------------
----
1F1t()–[] 1
2
---
1F1t()–[] 1
2
---1F2t()–[] 1
2
---
+
-----------------------------------------------------------------------=
1F1t()–
2F1t()–F2t()–
-----------------------------------------=
F
1
F2
,
Perror
()
Poutput 1 and input = 0=
()
=
Poutput 0 and input = 1=()+
Poutput 1 input = 0=()Pinput 0=()=
Poutput 0 input = 1=()Pinput 1=()+
ε1pε21p–()+=
Perror
()
Poutput 1= input = 0
()
=
Poutput 0= input = 1()+
pf2

Partial preview of the text

Download EECS 126 Midterm 1 Solutions: Probability Concepts and more Exams Probability and Statistics in PDF only on Docsity!

UNIVERSITY OF CALIFORNIA College of Engineering Department of Electrical Engineering and Computer Sciences

Professor Tse Fall 1998

EECS 126 — MIDTERM #1 Solutions

1a. (i) independent.

(ii) mutually exclusive.

(iii).

( reads, “If event occurs, then event must occur.)

(iv).

b. since.

2. Let be the lifetime of the machine picked and event that machine is picked.

Note: not necessarily exponential, as some of you have assumed.

3a.

Note: Some of you wrote:

a NO-NO!

E F , ⇒ P E F ( ) = P E ( ) = 0.

E F , ⇒ P E F ( )

P E ( ∩ F )

P F ( )

P ( E F )

P E ( ∩ F )

P F ( )

P F ( )

P F ( )

F ⊂ E F E

P ( E F )

P E ( ∩ F )

P F ( )

P E ( )

P F ( )

P F ( )

P ( B A )

P A B ( ) P B ( )

P A ( )

= -------------------------------- > P B ( ) P A B ( ) > P A ( )

X Ai = i

P A ( 1 X ≥ t )

P X ( ≥ t A 1 ) P A ( 1 ) P X ( ≥ t )

P X ( ≥ t A 1 ) P A ( 1 ) P X ( ≥ t A 1 ) P A ( 1 ) + P X ( ≥ t A 2 ) P A ( 2 )

[ 1 – F 1 ( ) t ]

[ 1 – F 1 ( ) t ] 1 2

--- [ 1 – F 2 ( ) t ] 1 2

1 – F 1 ( ) t 2 – F 1 ( ) t – F 2 ( ) t

F 1 , F 2

P ( error) = P ( output =1 and input = 0)

  • P ( output =0 and input = 1)

= P ( output = 1 input = 0) P ( input = 0 )

  • P ( output = 0 input = 1) P ( input = 1 )

= ε 1 p +ε 2 ( 1 – p )

P ( error) = P ( output = 1 input = 0)

  • P ( output = 0 input = 1)

b. Let be event that the bit gets flipped, ; let be event that we get a tail, , where are independent.

In part (a),

, for.

So the random rule is worse.

A p A ( ) = ε B p B ( ) = 1 – ε A B ,

P ( error) = P A ( ) P B ( ) + P A ( C ) P B ( C )

= ε ( 1 – ε) +( 1 – ε)ε

= 2 ε ( 1 – ε)

P ( error) = ε < 2 ε ( 1 – ε) ε