the big picture:
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The Big Picture:. Counting events in a sample space allows us to calculate probabilities The key to calculating the probabilities of events is to count the occurrences of events Our goal is to calculate probabilities of ALL possible events - PowerPoint PPT PresentationTRANSCRIPT
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The Big Picture:
• Counting events in a sample space allows us to calculate probabilities• The key to calculating the probabilities of
events is to count the occurrences of events• Our goal is to calculate probabilities of ALL
possible events• We can do this by counting the total number of
events possible in the sample space, using combinatorial mathematics to do this
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Permutations Permutation is a sequence or ordering of
events. Basic Question: if I have N objects, how
many different orderings of them are there? N! Formula: N(N-1)(N-2)…(1) Example: 5(5-1)(5-2)(5-3)(5-4)
5*4*3*2*1 = 120
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Permutations General formula for finding the number of
permutations of size k taken from n objects
PkN
N!
(N k)!
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Example
e.g. 10 compact discs and a 6 disc carousel- What is the number 6 disc orderings that we can make from the 10 cds?:
10! = 10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
(10 - 6)! 4! 4 x 3 x 2 x 1
= 10 x 9 x 8 x 7 x 6 x 5 = 151200
PkN
N!
(N k)!
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Combinations
General formula for finding the number of unique combinations of k objects you can choose from a set of n objects
CkN
N!
k!(N k)!
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Example
e.g. How many sets of 6 discs can we make from 10 cds (without repeating the same combinations)?
10! = 10! =10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 6!(10 - 6)! 6!(4!) (6 x 5 x 4 x 3 x 2 x 1)( 4 x 3 x 2 x 1)
= 10 x 9 x 8 x 7 = 5040 = 210(4 x 3 x 2 x 1) 24
CkN
N!
k!(N k)!
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Note the distinction Before we were counting our cd groupings without
regard to the same groupings of a different order But when looking for combinations, similar
groupings are not counted as different Pavement, The New Year, Built to Spill, Pixies, Grifters,
Sonic Youth Built to Spill, Grifters, Pavement, Sonic Youth, The New
Year, Pixies 2 permutations, but 1 combination (same set of 6 are involved)
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Bernoulli Trials Bernoulli trials = 2 mutually exclusive
outcomes Distribution of outcomes
e.g. the probability of various outcomes in terms of numbers of heads and tails
Order of items does not matter
N = # trials = 3
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Coin toss How many possible outcomes of the 3 coin tosses
are there? List them out:
HHH HHT HTT TTT TTH THH THT HTH Now condense them ignoring order
i.e. HTT = THT = TTH Only one occurrence of heads on 3 trials
Probability of 0 heads, only 1 heads, 2 heads, all 3 heads?
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Distribution of outcomes
Number of Heads
3.02.01.00.0
COIN FLIPSF
req
ue
ncy
4.0
3.0
2.0
1.0
0.0
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Now how about 10 coin flips? That’d be a lot of work writing out all the possibilities. What’s another way to find the probability of coin flips?
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Binomial distribution
Find a probability for an event using:
N = number of trials r = number of ‘successes’ p = probability of ‘success’ on any trial q = 1-p (probability of ‘failure’) CN
r=The number of combinations of N things taken r at a time
( ) ( )!( )
!( )!N r N r r N rr
Np r C p q p q
r N r
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So if I want to know the odds of getting 9 heads out of 10 coin flips or p(H,H, H,H, H,H, H,H, H,T):
p(9) =
10(.001953)(.5)=.0098 = .01
9 10 910!( ) ( )
9!(10 9)!p H p T
9 110!(.5) (.5)
9!(10 9)!9 110*9!
(.5) (.5)9!1!
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• Now if we did this for all possible hits (heads) on 10 flips:
Number Heads Probability (p value)
0 .001
1 .010
2 .044
3 .117
4 .205
5 .246
6 .205
7 .117
8 .044
9 .010
10 .001
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Using these probabilities
What is the probability of getting 4 or fewer heads in 10 coin tosses?
Addition
p(4 or less) = p(4) + p(3) + p(2) + p(1) + p(0) = .205 + .117 + .044 +.010
+ .001 = p = .377
About 38% of the time
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Now take it out a step Suppose you were giving some sort of treatment to
depressed individuals and assumed the treatment in general would have a 50/50 chance of doing so if it wasn’t anything special (i.e. just a placebo). Then it worked an average of 9 times out of 10 administrations.
Would you think there was something special going on or just chance?
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Not just 50/50 Not every 2 outcome
situation has equal probabilities associated with each option
There are two parameters we are concerned with when considering a binomial distribution. p = the probability of a
success. n = the number of Bernoulli
trials
Number of Successes
10987654321
Pro
ba
bili
ty
.4
.3
.2
.1
0.0
p = .8 N = 10 coins
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Also... More info about binomial distribution
= Np
2=Nqp
In Excel
=BINOMDIST(success,total N, prob., FALSE)
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Binomial distribution shapeApproximately “normal” curve when: p is close to 0.5
If not then “skewed” distribution N large
If not then not a representative distribution
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Binomial in Action Say someone claims ESP and we give them a test. The card can only
be a star or a circle on the one side and we let them guess as to what it is. We test them 20 times and they guess right on 13, which chance alone would dictate only 10 or .5 of the time.
2 Hypes: H0 = their performance is really no different from chance H1 = their performance is above chance.
The probability of guessing 13 is ~.07. Pretty unlikely!
However our question isn’t really about guessing 13 is it? It is about guessing 13 or more. Guessing at least 13 right has a chance of (or p value) ~.13 Still not that likely but not unreasonable. Think if there was a 13% chance of rain. We’d be surprised but not completely amazed.
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The problem with probabilities Can be very hard to grasp
e.g. Monty Hall problem TV show “Let’s make a deal” 3 closed doors, behind 1 is a prize Select a door Monty Hall opens one of the remaining doors
that does NOT contain a prize Now allowed to keep your original door or
switch to the other one Does it make a difference if you switch?