Transcript Lecture Notes

Psych 5500/6500
Probability
Fall, 2008
Gambler’s Fallacy
The “gambler’s fallacy” is to think that the “law
of averages” makes independent events no
longer independent. For example, on
flipping a fair coin, to think that if you get five
heads in a row then the probability of getting
a tail next is greater due to the ‘law of
averages’ is to make the gamblers fallacy.
Basics
1) p(A) = probability of event ‘A’
occurring
2) 0  p(A)  1.00
(p=0 means that A is an ‘impossible’ event,
p=1.00 means that A is a ‘sure’ event, probability
cannot be less than 0 or greater than 1.00)
3) p(~A) stands for probability of ‘not A’
4) p(A) + p(~A) = 1.00
(this is because A and ~A are mutually exclusive
and exhaustive)
Alternative Viewpoint
For an interesting alternative viewpoint on the
usefulness of dividing the world into A and ~A, see:
Kosko, B. (1993) Fuzzy Thinking: The New Science
of Fuzzy Logic. New York: Hyperion.
...but we digress
Theoretical Probability
If every outcome in a population of outcomes
has an equal chance of occurring, then:
the number of outcomes that fit ' A'
p(A) 
total number of possible outcomes
Examples
In a deck of 52 shuffled playing cards, the probability of drawing a
‘2’ is:
4
p(2) 
 0.0769
52
You have a bag of marbles containing 10 white, 5 red, 10 blue, and
25 yellow marbles, the probability of drawing a white marble would
be:
10
p(white) 
 .20
50
This formula requires that each outcome have an equal chance of
occurring (e.g. every card and marble have an equal chance of being
selected).
Empirical Probability
If you sample many times (with replacement)
from a population of outcomes, then:
number of times ' A' occurred while sampling
p(A) 
number of times you sampled
This assumes each sample event is independent. You need to
sample ‘with replacement’ so that you are not drawing from a
different population each time you sample.
Example
You have a bag of marbles of various colors. You
want to know the probability of drawing a green
marble. You sample 50 times, each time you draw
a marble, note its color, replace the marble in the
bag, and mix the marbles.
Results: 20 green, 10 red, 5 white, 15 blue marbles
during 50 draws.
20
p(drawing a green marble) 
 .40
50
Note that with this formula every outcome (i.e. marble) does not
have to have an equal chance of occurring.
Examples
Of the 134 times you’ve dropped a peanut butter sandwich while in
the process of making it, it has landed peanut butter side down 99
times.
99
p(peanut butter side down) 
 0.74
134
Note that this formula assumes that each draw is made from the
same population, in this case that would require that you do not
change your behavior over trials due to practice or frustration.
Conditional Probability
p(A|B) = the probability of Event A given
Event B. B is the ‘condition’ in which the
probability of A is being determined.
Examples
p(A|B)
The probability of a card being a diamond
given that it is red. ‘A’=diamond, ‘B’=red.
The probability of improvement given the
subject was in the ‘treatment’ group. ‘A’ =
improvement, ‘B’ = was in the treatment
group.
Conditional Probability (cont.)
p(A|B) + p (~A|B) = 1.00
‘A’ = catching a cold
‘B’ = getting your hair wet
The probability of catching a cold given your hair got wet,
plus the probability of not catching a cold given you hair
got wet = 1.00 (i.e. if your hair gets wet you will either
catch a cold or not)
p(A|B) + p(A|~B) does not necessarily equal 1.00.
The probability of a catching a cold given your hair got wet,
plus the probability of catching a cold given your hair
didn’t get wet = ?
Conditional Probability and
Independence of Events
If p(A|B) = p(A) then events A and B are
independent. Note, if the above is true then
it will also be found that p(B|A)=p(B);
Example: in drawing from a shuffled, standard
deck of cards. ‘A’ is ‘drawing a 2’, ‘B’ is
‘drawing a heart’. p(A|B): the probability of
the card being a 2 given that it is a heart =
p(A) the probability of drawing a 2. B does
not change the probability of A, thus ‘suit’
and ‘value’ are independent.
Confusing p(A|B) and p(B|A)
A common mistake is to think that p(A|B) =
p(B|A)
This may or not may seem that common,
wait until we talk about null hypothesis
testing!!! Many people make this mistake.
Example p(A|B)  p(B|A)
p(card being a heart | the card was red) = 0.5
p(card being red | the card was a heart) = 1.0
Conditional Probability: Bayes’s
Theorem
If you want to know the relationship between p(A|B) and the p(B|A)
then here it is (believe it or not this will be useful):
p(B | A)p(A)
p(A | B) 
p(B | A)p(A)  p(B |~ A)p(~ A)
Probability and the Normal
Distribution
If we know that 21% of a population (a
proportion of 0.21) is left-handed, and we
randomly sample one person from that
population, what is the probability that we
will sample someone who is left-handed?
p=0.21 of course.
– Keep that in mind...
Normal Distribution Example 1
We are going to randomly sample from a
population that is normally distributed with a
mean of 120 and a standard deviation of 28.
What is the probability of selecting a person
with a score that is at least 15 below the
mean?
p(A) = p(Y105) = ?
Draw a normal distribution representing the population, label it
as such, label the mean and standard deviation on the
curve, shade in the area that fits event A, compute the value
of z, look up what proportion of the curve falls in that area.
Probability
If .2946 of the population has a score of 105
or less, then the probability of randomly
selecting one person with a score of 105 or
less is also .2946.
p(Y105) = .2946 (or 29.46%)
Normal Distribution Example 2
We are going to randomly sample from a
population that is normally distributed with a
mean of 50 and a standard deviation of 16.
What is the probability of selecting a person
with a score that is within eight of the mean?
p(A) = p(42  Y  58) = ?
Draw a normal distribution representing the population, label it
as such, label the mean and standard deviation on the
curve, shade in the area that fits event A, compute z’s, look
them up in the table.
Original Population
p(42  Y  58) =.1915 + .1915 = .3830 or 38.30%
One-tailed and two-tailed p
values
We are going to want to ask questions that go
in the opposite direction. For example, what
value of z cuts off .05 (i.e. 5%) of the upper
side of the curve (i.e. the value of z that
there is a 5% chance of equaling or
exceeding). The probabilities we are going
to be the most interested in are .05 and the
.01
What value of z accomplishes this?
Getting the z for p
To answer this we look at the table of z
values, first finding the probability we want
(p=.05) and then looking to see what value
of z cuts off that proportion of the curve. If
we do we find that the value of z would be
between z=1.64 and z=1.65. If we use the
‘Normal Distribution Tool’ I wrote then we
find that the value of z is 1.645. We will
round that off to 1.65
There is a 5% chance of sampling someone with a z score of 1.65 or greater
There is a 5% chance of sampling someone with a z score of -1.65 or less
There is a 5% chance of sampling someone with a z score of –1.96 or less, or a
z score of 1.96 or more.
There is a 1% chance of sampling someone with a z score of 2.33 or greater
There is a 1% chance of sampling someone with a z score of -2.33 or less
There is a 1% chance of sampling someone with a z score of -2.58 or less, or a
z score of 2.58 or greater.