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03-Conditional Prob

# 03-Conditional Prob - IE 111 Fall Semester 2009 Note Set#3...

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IE 111 Fall Semester 2009 Note Set #3 9/7/09 Conditional Probability and Independence Conditional Probability and Independence It is often useful to be able to calculate the probability of some event given that some other event has occurred. Example Experiment: Flip 3 coins. Event A = "2 or more heads" Event B = "first coin is head" We can ask, what is Pr{ 2 or more H given the first coin is a head } Or in mathematical notation: P(A|B) We would use the following notation in this example: Pr{A|B} where the "|" is always read "given" or "given that". Also note that "|" is different from "/" and does not have anything to do with dividing. The following is very important: Definition of Conditional Probability ) ( ) ( ) | ( B P B A P B A P = The above equation is the definition of conditional probability . It defines rigorously the term "given that" and the symbol "|". Note that we assume P(B) > 0 since the "given B" part would make no sense if it had zero probability. That is, if P(B)=0 it could not be the case that we are "given that B occurred". Next: a couple of simple examples. Example 1 Returning to the example above: Flip 3 coins. 1

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A = 2 or more heads B = first coin is head What is Pr{A|B} A = {HHT, HTH, THH, HHH} so P(A) = 4/8 = 0.5 B = {HHH, HHT, HTH, HTT} so P(B) = 4/8 = 0.5 {A and B} = {HHH, HHT, HTH} so P(A and B) = 3/8 Thus P(A|B) = (3/8) / (4/8) = 3/4 = 0.75 Note that P(A|B) is different from P(A). Knowing that B occurred changes our assessment of the probability of A. In this case P(A|B) > P(A). This makes sense intuitively. Knowing for sure that the first coin is a head would seem to increase the chances of 2 or more heads. Example 2 Suppose the class breaks down this way: Male Female IE 11 7 MATH 5 3 CHE 4 1 Thus we have 31 students; 20 men and 11 women; 18 IE’s, 8 MATS, and 5 ChE’s Suppose we select a student at random. Find the following using the definition of conditional probability. a) P(IE) = 18/31 b) P(female) = 11/31 c) P(IE and Female) = 7/31 d) P(MATS and Female) = 3/31 e) P(IE|Female) = (7/31) / (11/31) = 7/11 f) P(Female|IE) = (7/31) / (18/31) = 7/18 2
g) P(Male and Female) = 0 because these are mutually exclusive h) Is it always true that P(B|A) = P(A|B) ?

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