16.322 Stochastic Estimation and Control
Professor Vander Velde
Lecture 2
Lecture 2
Last time:
Given a set of events with are
mutually exclusive
and
equally likely
,
()
nE
PE
N
=
.
Example: card games
Number of different 5card poker hands
52
2,598,960
5
⎛⎞
==
⎜⎟
⎝⎠
Number of different 13card bridge hands
52
635,013,559,600
13
Poker probabilities:
1.
4444
13
12
11
10
2111
(one pair)
0.423
52
3!
5
P
⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞
⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟
⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠
2.
444
13
12
11
221
(two pair)
0.0476
52
2!
5
P
⎛⎞ ⎛⎞ ⎛⎞
⎜⎟ ⎜⎟ ⎜⎟
⎝⎠ ⎝⎠ ⎝⎠
Independence
(
)
(
)
PAB PAPB
=
when A,B independent
(
)
(

)
PAB PAPB A
=
when A,B dependent
Definition of independence
of events:
123
1
2
(
...
)
(
) (
)... (
)
mm
PAAA A
PAPA PA
=
for all m
(pairwise, threewise, etc.)
Definition of conditional probability
:
()
PAB
PB A
PA
=
restricts our attention to the situation where A has already occurred.
If A,B are independent
()()
PAPB
PB
Two useful results:
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View Full Document16.322 Stochastic Estimation and Control
Professor Vander Velde
Lecture 2
1.
(
...)
( ) (

) (

) (

)...
P ABCD
P A P B A P C AB P D ABC
=
Derive this by letting
A=CD.
Then
()
(
)
(

)
(
)
(

)
(

)
PBCD PCDPBCD PCPDCPDCD
==
2.
If
1
A
,
2
A
,… is a set of
mutually exclusive
and
collectively exhaustive
events, then
12
1
1
2
2
( )
(
)
(
)
...
(
)
(
) (

)
(
) (

)
...
(
) (

)
nn
n
PE
PEA
PAPE A
PA PE A
=
+
++
=
+
(
)
(
)
(
)
PA B PA PB PAB
+=
+
−
must subtract off P(AB) because it is counted
twice by the first two terms of the RHS.
(
)
(
)
(
)
(
)
(
)
(
)
(
)
PA B C PA PB PC PAB PAC PBC PABC
++ =
+
+
−
−
−
+
(for three
events).
If four events,
PA B C D
+++
would be the sum of probabilities of the single
events, minus the probability of the joint events taken two at a time, plus the
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 Fall '04
 WallaceVanderVelde
 Probability, Probability theory, Professor Vander Velde, Control Professor Vander

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