# It is then natural to consider the events to be

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Unformatted text preview: 36 = 6 . Since EF = {(2, 5), (4, 3), (6, 1)}, P (EF ) = P (EF ) 1/12 3 1 1 1 36 = 12 . Since E has 18 outcomes, P (E ) = 2 . Therefore, P (F |E ) = P (E ) = 1/2 = 6 . Note that, for this example, P (F ) = P (F |E ). That is, if one learns that the number coming up on the ﬁrst die is even, the conditional probability that the numbers coming up on the dice sum to seven is the same as the original probability that the numbers sum to seven. We comment brieﬂy on some properties of conditional probabilities. These properties follow easily from the deﬁnition of conditional probabilities, and the axioms of probability. Suppose A is an event with P (A) > 0, and B is another event. Then 1. P (B |A) ≥ 0. 2. P (B |A)+ P (B c |A) = 1. More generally, if E1 , E2 , . . . are disjoint events, P (E1 ∪ E2 ∪· · · |B ) = P (E1 |B ) + P (E2 |B ) + · · · . 3. P (Ω|B ) = 1. 4. P (AB ) = P (A)P (B |A). 30 CHAPTER 2. DISCRETE-TYPE RANDOM VARIABLES 5. P (ABC ) = P (C )P (B |C )P (A|BC ) (assuming P (BC ) > 0.) The ﬁrst three properties above are equi...
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## This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Institute of Technology.

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