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09-28 Conditional Probability and Independence (2)

# 09-28 Conditional Probability and Independence (2) - BTRY...

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BTRY 4080 / STSCI 4080 Fall 2010 145 The Bayes Formula For events E S and F S such that P ( E ) > 0 , P ( F | E ) = P ( E | F ) P ( F ) P ( E | F ) P ( F ) + P ( E | F c ) P ( F c ) Proof: Using the definition of conditional probability: P ( F | E ) = P ( EF ) P ( E ) = P ( E | F ) P ( F ) P ( E ) .

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BTRY 4080 / STSCI 4080 Fall 2010 146 Now, write S = F F c . Then, notice that P ( E ) = P ( ES ) = P ( EF )+ P ( EF c ) = P ( E | F ) P ( F )+ P ( E | F c ) P ( F c ) Hence, P ( F | E ) = P ( E | F ) P ( F ) P ( E ) = P ( E | F ) P ( F ) P ( E | F ) P ( F ) + P ( E | F c ) P ( F c ) . square
BTRY 4080 / STSCI 4080 Fall 2010 147 Example (3.3.3a): An accident-prone person will have an accident within a fixed one-year period with probability 0 . 4 . This probability decreases to 0 . 2 for a non-accident-prone person. Suppose that the probability of a randomly selected person being accident prone is 0 . 3 (i.e., in the population, 30% are accident-prone). Let A = { accident prone } and B = { has accident } . Problem specifies that P ( B | A ) = 0 . 4 , P ( B | A c ) = 0 . 6 , P ( A ) = 0 . 3 . Q 1 : What is the probability that a randomly selected person will have an accident within a fixed one-year period? That is: P ( B ) ? Q 2 : Suppose a person has an accident within a year. What is the probability that he/she is accident prone? That is: P ( A | B ) ?

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BTRY 4080 / STSCI 4080 Fall 2010 148 Example (3.3.3c): Consider a multiple choice test question. Let p
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