L16_Independence_print

# L16_Independence_print - 1 COMP170 Discrete Mathematical...

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Unformatted text preview: 1 COMP170 Discrete Mathematical Tools for Computer Science Discrete Math for Computer Science K. Bogart, C. Stein and R.L. Drysdale Section 5.3, pp. 236-247 Independence Version 2.0: Last updated, May 13, 2007 Slides c 2005 by M. J. Golin and G. Trippen 2 Conditional Probability and Independence • Conditional Probability • Independent Trials Processes • Independence 3 Conditional Probability Suppose we’ve thrown two fair dice. The probability of seeing “double-twos” is 1 36 . Now suppose that we don’t see the dice but know that the event “the dice sum up to 4 ” has occured. What is the probability that “double-twos” occurred given that “the dice sum up to 4 ”? { , , Answer “should be” 1 3 , shouldn’t it? This lecture formalizes this intuition. 4 2 dice: Event ”at least one circle on top” is: { } , , Applying principle of inclusion and exclusion : probability of seeing a circle on at least one top when we roll the dice is 1 3 + 1 3- 1 9 = 5 9 A more complicated example , , 5 Suppose you are told that the two dice have been rolled and both top shapes are the same ? What is the probability that at least one top shape (and now therefore both top shapes) is a circle? P ( circles ) = 4 P ( triangles ) Originally, (i) chance of getting (two) circles was 4 times chance of getting (two) triangles and (ii) chance of getting (two) squares was 9 times chance of getting (two) triangles so, given that both top shapes are the same intuitively, we “should ”have and P ( squares ) = 9 P ( triangles ) 6 Let p = P ( triangles ) . Since probabilities sum to 1 , P ( circles ) = 4 P ( triangles ) and P ( squares ) = 9 P ( triangles ) P ( two circles if both tops are the same ) = 4 p = 2 7 . p + 4 p + 9 p = 1 or p = 1 14 , and 7 Do these analyses make sense? How can we replace intuitive calculations with a formula that we can use in similar situations? WARNING There are situations where our intuitive idea of probability does not always agree with what the rules of probability give us! 8 { TT, TC, TS, CT, CC, CS, ST, SC, SS } . 1 36 1 18 1 12 1 18 1 9 1 6 1 12 1 6 Rolling our two unusual dice Original sample space with probabilities 1 4 We know that event { TT, CC, SS } happened. Thus, although this event used to have probability 1 36 + 1 9 + 1 4 = 14 36 = 7 18 it now has probability 1 . Given this, what probability should we assign event of seeing a circle ( CC ) ? 9 Multiply all three old probabilities by 18 / 7 : new probabilities will preserve ratios and sum to 1 . P ( two circles ) = 1 9 · 18 7 = 2 7 We now capture this reasoning process in a formula! { TT, CC, SS } New Sample Space 1 36 1 9 1 4 Probabilities in old sample space Sum is 7 18 New Probabilities 1 14 2 7 9 14 Sum is 1 10 The conditional probability of E given F , denoted by P ( E | F ) (read as ”the probability of E given F”) is P ( E | F ) = P ( E ∩ F ) P ( F ) ....
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L16_Independence_print - 1 COMP170 Discrete Mathematical...

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