(3) By symmetry, then, P(A|B) = P(A).
From the intuitive idea, it seems clear that
if A and B are independent then so are Acand B,
A and Bc , and Acand Bc
(Proofs: Handout)
Then P(B|Ac) = P(B) = P(B|A), the
rest of the intuitive idea.
For constructing models: we think not
just of events being independent, but whole mechanisms or processes.
Every event in process A being independent of any event in process B. (e.g.
Flip a coin, then draw a card from a deck)
Trees: If two processes are independent,
and can be thought of as sequential in a tree, all the branchings at the
second stage will be identical, whatever the preceding result.
Homework:
Handout--show: If A and B are independent,
then so are Acand B, A and Bc ,
and Acand Bc
Ash p. 45: 8,
11.
12 (is it the same for W on the
1st and the 2nd? See "time blindness" , Monday)
M&M p. 310 ff: 4.28, 29,
30, 32, 34, 36 (Some may be repeats from last semester. Solutions
are in the Math Clinic)
What's Next? , review M&M pp. 312-317
(Random variables), Ash sec. 2.2 Multinomial and Binomial distributions.
(Ash does Multinomial first, and Binomial as a special case. This
approach doesn't "privilege" the binomial "successes" as much as M&M's
approach did.)
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