HW: One
more conditional probability, shouldn't be hard: M&M p.330-1, 4.124
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 (2nd and 3rd can be anything. Is it
the same answer as for W on the 1st and the 2nd? See "time
blindness", Day 7 . If you don't buy
the time blindness, do a tree
and calculate the answer from it.)
M&M 5th ed. p274ff. 4.23, 4.28,
4.30, 4.31, 4.27, 4.33, 4.37 and 39. also p.330 4.123 (Some may be repeats from last
semester.)
= = = = = = = = = = = = = = = = = = = = = =
(1) Definition: A and B are
independent
iff P(A and B) = P(A)·P(B)
Intuitive:
Event A and event B are independent if the occurrence or nonoccurence
of
one makes no difference to the probability of the other.
(2) Formula: P(B|A)
= P(B)
(1) and (2) are (almost) equivalent,
since
P(B|A) = P(BandA)/P(A). (proofs, in class)
(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.
That was about independence of events.
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.
What's Next? , review M&M pp. 277-286 (4.3 Random variables), pp. 335-37 & 348-350 (Binomial). 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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