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| + | [[Category:ECE302Fall2008_ProfSanghavi]] | ||
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=Independence= | =Independence= | ||
==In Two Events== | ==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> | <math>P(A \bigcap B) = P(A) \times P(B) </math> | ||
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For example, given a coin, are the two outcomes independent? | For example, given a coin, are the two outcomes independent? | ||
| − | <math> P( \lbrace C_1=H \rbrace \bigcap \lbrace C_2 =H \rbrace )</math> | + | <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)</math> | + | <math> P( C_1=H ) \times P(C_2=H) = 1/2 \times 1/2 = 1/4</math> |
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| + | Since the product of the two probabilities is equal to overall probability, the events are independent. | ||
[http://en.wikipedia.org/wiki/Help:Formula] | [http://en.wikipedia.org/wiki/Help:Formula] | ||
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==In Multiple Events== | ==In Multiple Events== | ||
| − | <math> \ | + | <math> P( \bigcap_{i \in S} A_i ) = \prod_{i \in S} P(A_i)</math> for all sets <math> S </math> of events. |
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==Conditional Probability== | ==Conditional Probability== | ||
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<math>P(A \bigcap B|C) = P(A|C) \times P(B|C)</math> | <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.
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) $
