(New page: ==Independence in Probability== ==Independence in Multiple Events== ==Independence in Conditional Probability==)
 
 
(16 intermediate revisions by 4 users not shown)
Line 1: Line 1:
==Independence in Probability==
+
[[Category:ECE302Fall2008_ProfSanghavi]]
 +
[[Category:probabilities]]
 +
[[Category:ECE302]]
  
==Independence in Multiple Events==
+
=Independence=
  
==Independence in Conditional Probability==
+
==In Two Events==
 +
Two events A and B are independent if the following formula holds:
 +
 
 +
<math>P(A \bigcap B) = P(A) \times P(B) </math>
 +
 
 +
For example, given a coin, are the two outcomes independent?
 +
 
 +
<math> P( \lbrace C_1=H \rbrace  \bigcap  \lbrace C_2 =H \rbrace  ) = 1/4</math>
 +
 
 +
<math> P( C_1=H ) \times P(C_2=H) = 1/2 \times 1/2 = 1/4</math>
 +
 
 +
Since the product of the two probabilities is equal to overall probability, the events are independent.
 +
 
 +
[http://en.wikipedia.org/wiki/Help:Formula]
 +
 
 +
==In Multiple Events==
 +
 
 +
<math> P( \bigcap_{i \in S} A_i ) = \prod_{i \in S} P(A_i)</math> for all sets <math> S </math> of events.
 +
 
 +
==Conditional Probability==
 +
 
 +
A & B are conditionally independent given C if the following formula holds true.
 +
 
 +
<math>P(A \bigcap B|C) = P(A|C) \times P(B|C)</math>
 +
----
 +
[[Main_Page_ECE302Fall2008sanghavi|Back to ECE302 Fall 2008 Prof. Sanghavi]]

Latest revision as of 12:00, 22 November 2011


Independence

In Two Events

Two events A and B are independent if the following formula holds:

$ P(A \bigcap B) = P(A) \times P(B) $

For example, given a coin, are the two outcomes independent?

$ P( \lbrace C_1=H \rbrace \bigcap \lbrace C_2 =H \rbrace ) = 1/4 $

$ P( C_1=H ) \times P(C_2=H) = 1/2 \times 1/2 = 1/4 $

Since the product of the two probabilities is equal to overall probability, the events are independent.

[1]

In Multiple Events

$ P( \bigcap_{i \in S} A_i ) = \prod_{i \in S} P(A_i) $ for all sets $ S $ of events.

Conditional Probability

A & B are conditionally independent given C if the following formula holds true.

$ P(A \bigcap B|C) = P(A|C) \times P(B|C) $


Back to ECE302 Fall 2008 Prof. Sanghavi

Alumni Liaison

Followed her dream after having raised her family.

Ruth Enoch, PhD Mathematics