let’s talk about conditional probability by considering a specific example: –suppose we roll a pair of dice and are interested in the probability of getting an 8 or more (sum of the spots >= 8). what is the unconditional probability of this happening? –now what if when I roll the dice, one of them rolls under the chair, and all I can see is the other die with 5 spots on the up-face. what is the conditional probability that the sum of the spots >= 8 given that one of the dice has 5 spots? –notice that knowledge of the one die’s 5 spots essentially changes the sample space from S of 36 points to one of just 6 points: (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) and so the probability should be 4/6=2/3 –note that this coincides with the definition of conditional probability given on page 80: since P({(5,3),(5,4),(5,5),(5,6)} / P({(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)}) = (4/36) / (6/36) = 4/6 = 2/3 –we usually use this relationship to compute probabilities of non- independent events:

and then we define two events to be independent whenever then we get the usual formula for “and” for independent events: go over the water quality example on page 83. this example assumes that successive water samples are independent of each other... a commonly used application of conditional probability is given in Bayes Theorem (page 87) – but first let’s look at a preliminary result called the theorem of total probabilty via the example at the top of page 85: suppose a plant gets part VR from one of three suppliers (B 1, B 2, B 3 ): 60% of all VRs come from B 1, 30% come from B 2, and 10% come from B 3. the three suppliers have varying records as to the quality of their product: (95%, 80%, 65%) perform as specified. Choose a VR at random – what is the probability that it performs as specified? let A=“VR performs as specified”. then show the total probability of A as broken up into its “parts” as determined by the three suppliers (do a Venn diagram and a tree to show how this works...)

now consider a related problem that can be solved by Bayes Theorem: what is the probability that a randomly chosen VR is from supplier B 1 given that it performs to specifications? Notice that this is the reverse of the probabilities in the theorem of total probabilities...so first write then rewrite the denominator in terms of the theorem of total probability and we have Bayes Theorem (Theorem 3.11 on page 87) note that the numerator of Bayes Theorem is the probability of A going through the rth branch of the tree and the denominator is the sum of the probabilities of A going through all the branches of the tree... see the rest of the author’s notes on this theorem on page 87. go over the example at the bottom of page 87 HW: Read 3.6 and 3.7 and do the following problems:3.64, 3.66, 3.67, 3.69, , 3.79