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Probability

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Probability theory appeared between 1654 (a minute and a half ago) and 1843 (34 seconds ago) ... emergence of elementary probability theory in the 1650s met ... – PowerPoint PPT presentation

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Title: Probability


1
Probability Certainty
2
Overview
  • History of probability theory
  • Some basic probability theory
  • Calculating simple probabilities
  • Combining mutually-exclusive probabilities
  • Combining independent probabilities
  • More complex probabilities
  • Calculating conditional probabilities
  • Bayes Theorem, and why we should care about it
  • A test case The Lets Make A Deal! problem

3
History of probability theory
  • Compress all of human history (350K years) in one
    24-hour day
  • the first recorded general problem representation
    (geometry, invented by Thales of Miletus about
    450 B.C.) would have appeared only 9 minutes and
    30 seconds ago
  • the first systematic large-scale collection of
    empirical facts (Tycho Brahes collection of
    astronomical observations) would have appeared
    only a minute and a half ago
  • the first mathematical equation which was able to
    predict an empirical phenomena (Newtons 1697
    equation for planetary motion) would have
    appeared only one minute and twelve seconds ago
  • Probability theory appeared between 1654 (a
    minute and a half ago) and 1843 (34 seconds ago).

4
History of probability theory
  • The emergence of elementary probability theory in
    the 1650s met with enormous resistance and lack
    of comprehension when it was first introduced,
    despite its formal character, its utility, and
    (what we now recognize as) its simplicity.
  • The difficult points (Margolis, 1993) were
    philosophical rather than mathematical the
    notions that one could count possibilities that
    had never actually existed, and that order could
    be obtained from randomness.

5
Basic probability theory Example 1
  • A boring standard example

How likely is it that we will throw a 6 with one
dice?
  • Basic principle The probability of any
    particular event is equal to the ratio of the
    number of ways the event can happen over the
    number of ways anything can happen ( the number
    of ways the event can fail to happen the number
    of ways it can happen).

6
Basic probability theory Example 2
How likely is it that we will throw a 7 with two
die?
  • There is more than one way for the event to
    happen 6 1, 16, 5 2, 25, 3 4, 4 3 6
    ways
  • There are 36 (6 x 6) ways for anything to happen
  • So the probability of 7 with two die is 6/36 or
    1/6.

7
Basic probability theory Example 3
We roll the die 4 times, and never get a seven.
What is the probability that well get on the
5th roll?
  • Independent events are events that dont effect
    each others probability. Since the every roll is
    independent of every other, the odds are still
    1/6.

8
Basic probability theory Example 4
We roll the die twice and get a seven both times.
What is the probability of that?
  • To combine probabilities of independent events,
    multiply the odds of each event. The odds of each
    7 are 1/6, so the odds of two rolls of 7 in a row
    are 1/6 x 1/6 1/36.

9
Basic probability theory Example 5
We roll the die. What are the odds are getting
either a 7 or a 2?
  • Now the events are mutually exclusive if one
    happens, the other cannot. To combine
    probabilities of mutually exclusive events, add
    them together.
  • 6/36 odds of 7 1/36 odds of 2 7/36

10
Basic probability theory Example 6.1
We roll the die twice. What are the odds that we
get at least one 7 from the two rolls?
  • Can we just add the probabilities of getting a 7
    on each roll?
  • No because the events are not mutually
    exclusive anymore
  • if we could, wed be above 100- guaranteed a 7-
    after 6 rolls! Where could the guarantee come
    from?

11
Basic probability theory Example 6.2
We roll the die twice. What are the odds that we
get at least one 7 from the two rolls?
  • Can we just multiply the probabilities of
    getting a 7 on each roll?
  • No because the events are not independent (Why
    not?)
  • And 1/6 x 1/6 is less than 1/6- so wed have
    smaller chance with two rolls than with one!

12
Basic probability theory Example 6.3
We roll the die twice. What are the odds that we
get at least one 7 from the two rolls?
  • We can turn part of the problem into a problem
    of mutual exclusivity by asking what are the
    odds of there being exactly one seven out of two
    rolls?
  • one way is to roll 7 first, but not second
  • - the odds of this are 1/6 5/6 (independent
    events) 0.138
  • - the odds of rolling 7 second are 5/6 1/6
    (independent events) 0.138
  • - since these two outcomes are mutually
    exclusive, we can add them to get 0.138 0.138
    0.277

13
Basic probability theory Example 6.4
We roll the die twice. What are the odds that we
get at least one 7 from the two rolls?
  • Now we need to know what are the odds of there
    being two sevens out of two rolls?
  • We already know its 1/36 0.03
  • So the odds that we get at least one 7 is the
    odds of two 7s the odds of one 7 (mutually
    exclusive events) 0.03 0.27 0.3

14
Basic probability theory The generalization
  • Does anyone know how to generalize this
    calculation, so we can easily calculate the odds
    of an event of probability p happening r times
    out of n tries, for any values of p,r, and n?
  • The binomial theorem Tune in for the exciting
    next class!

15
Basic probability theory Conditional probability
  • What if a dice is biased so that it rolls 6
    twice as often as every other number? How can we
    deal with uneven base rates?
  • Why should we care?
  • Because real life uses biased dice
  • eg. the conditional probability of being
    schizophrenic, given that a person has an
    appointment with a doctor who specializes in
    schizophrenia, is quite different from the
    unconditional probability that a person has
    schizophrenia (the base rate)

16
Basic probability theory Conditional probability
  • A particular disorder has a base rate occurrence
    of 1/1000 people. A test to detect this disease
    has a false positive rate of 5. Assume that the
    test diagnoses correctly every person who has the
    disease. What is the chance that a randomly
    selected person with a positive result actually
    has the disease? Take a guess.

Harvard Med School estimates About half said
95. Average response was 56. Only 16 gave the
correct answer.
17
Basic probability theory Conditional probability
  • A particular disorder has a base rate occurrence
    of 1/1000 people. A test to detect this disease
    has a false positive rate of 5. Assume that the
    test diagnoses correctly every person who has the
    disease. What is the chance that a randomly
    selected person with a positive result actually
    has the disease?
  • Conditional probability P(AB) P(A,B) / P(B)
    P(A and B)/P(B)
  • Let A Has the disorder and B Has a positive
    test result
  • In 10,000 people, P(A and B)
  • P(B)
  • Chance that a randomly selected person with a
    positive result actually has the disease

0.001 1000 10
0.05 1000 (false positives) 10 (true
positives) 510
10/510 0.02 or just a 2 chance
18
A generalization Bayes theorem
P(AB) P(BA) P(A) / P(B) Proof By
definition, (1.) P(AB) P(A,B) / P(B) (2.)
P(BA) P(A,B) / P(A) (3.) P(BA) P(A) P(A,B)
Multiply (2.) by P(A) (4.) P(AB) P(B) P(BA)
P(A) Substitute (1.) in (3.) (5.) P(AB)
P(BA) P(A) / P(B)
19
Lets try Bayes theorem
P(AB) P(BA) P(A) / P(B) Let P(A)
Probability of disease 0.001 P(B)
Probability of positive test 0.05 0.001
0.0501 P(BA) Probability of positive
test given disease 1 Then P(AB) P(BA)
P(A) / P(B) (1 x
0.001) / (0.0501) 0.02
20
The Notorious 3-Curtain (Lets Make A Deal)
Problem!
  • Three curtains hide prizes. One is good. Two are
    not.
  • You choose a curtain.
  • The MC opens another curtain. Its not good.
  • He gives you the chance to stay with our first
    choice, or switch to the remaining unopened
    curtain.
  • Should you stay or switch, or does it matter?

21
  • P(AB) P(BA) P(A) / P(B)
  • Assume you choose A and switch to the unopened
    one
  • LetP(A) Probability it is behind the unopened
    one (B or C) 2/3
  • P(B) Probability it is not behind A 2/3
  • P(BA) Probability it is not behind A, given
    that it is behind the unopened one (B or C) 1
  • Then P(AB) Probability it is behind the
    unopened one (B or C), given that it is not
    behind A
  • P(BA) P(A) / P(B)
  • (1 x 2/3) / 2/3 1 100 of
    the time.
  • But it is given that it is not behind A 2/3 of
    the time, so 2/3 of the time we can be certain of
    winning if we switch!
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