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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:
P( A and B) P( A  B)
P( A | B) 

P( B)
P( B)
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 nonindependent events:
P( A and B)  P( A) P( B | A)  P( B) P( A | B)
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and then we define two events to be independent whenever
P( B | A)  P( B) (or P( A | B)  P( A))
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then we get the usual formula for “and” for independent events:
P( A and B)  P( A) P( B)
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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 (B1, B2, B3): 60% of
all VRs come from B1, 30% come from B2, and 10% come from B3. 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 B1 given that it performs to specifications? Notice that this
is the reverse of the probabilities in the theorem of total
probabilities...so first write
P( B1 and A) P( B1 ) P( A | B1 )
P( B1 | A) 

P( A)
P( A)
• 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, 3.71-3.75, 3.79