This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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. 236247 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 “doubletwos” 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 “doubletwos” 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 ) ....
View
Full Document
 Spring '10
 M.J.Golin
 Computer Science, Conditional Probability, Probability, Probability theory, independent trials

Click to edit the document details