PROBABILITY

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PROBABILITY
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Simplest Notions of Probability -coin tossing
and dice rolling
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Coin Tossing
First toss
Second toss
HH
H
H
HT
T
TT
T
T
TH
H
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Coin Tossing
Probability of sequence
First toss
Second toss
HH
H
0.5 X 0.5 0.25
0.5
H
HT
T
0.5
0.5 X 0.5 0.25
0.5
TT
T
0.5 X 0.5 0.25
0.5
T
TH
H
0.5
0.5 X 0.5 0.25
0.5
ALWAYS!
S 1.0
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Tossing Dice
Question Whats the chance of getting a
combination of numbers on dice that adds up to 3?
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Question Whats the chance of getting a
combination of numbers on dice that adds up to 3?
Chance of any number showing up is 1/6 BUT There
is more than one way of getting a 3
or


1/6 x 1/6 1/36
1/6 x 1/6 1/36
Chance 1/36 1/36 1/18
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Question Whats the chance of getting a
combination of numbers on dice that adds up to 3?
-using a tree diagram as in the coin example
First toss
Second toss
Total of 2 dice
1 2 3 4 5 6
2 3 4 5 6 7
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1/6 chance
1/36 chance 0f each
1/6 chance of each
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Do tree diagrams for every possible combination
1
3
4
1
2
4
5
2
2
3
5
1/36 chance of each
6
3
3
1/36 chance of each
4
6
7
4
5
7
8
5
6
8
9
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And for 4, 5 and 6 as first toss
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End up with the following
Sum of two dice
Number of ways to achieve it (or probability of
getting it)
2 1/36 3 2/36 4 3/36 5 4/36 6 5/36 7 6
/36 8 5/36 9 4/36 10 3/36 11 2/36 12 1/3
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What about something biological?? Can this kind
of stuff be used In biological problems you might
encounter? Question If we have two alleles for
eye colour in Drosophila that behave in a
Mendelian fashion, what are the chances of all
the possible allele combinations?
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This is the kind of reasoning that lead to the
development of Punnett Squares and the
Hardy-Weinberg Equilibrium
Punnett
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F2
x
Chromosome with red allele
Chromosome with white allele (recessive)
If you look on reproduction as picking
chromosomes, then
First pick
Second pick
Probability of sequence
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F2
x
Chromosome with red allele
Chromosome with white allele (recessive)
If you look on reproduction as picking
chromosomes, then
First pick
Second pick
Probability of sequence
0.25
0.5
0.25
3 in 4 chances of red
0.5
0.5
0.25
0.5
0.5
0.25
1 in 4 chances of white
0.5
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Statistical Independence - The Promotion Problem
Question Does company X show sex discrimination
in promotion?
Sample of 200 employees Results 50 promoted
males 70 males - not promoted 15 promoted
females 65 females - not promoted
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Contingency Table
Are two events, sex and promotional status,
statistically independent?
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Now - compute a series of probabilities
Probability of selecting a female 80/200
.4 Probability of selecting a female 15/65
.23 (given that we select a promoted
individual) Probability of probability of being
promoted if you are male 50/120
.417 Probability of probability of being promoted
if you are female 15/80 .188 Overall
probability of being promoted 65/200 .325
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Now - compute a series of probabilities
Probability of selecting a female 80/200
.4 Probability of selecting a female 15/65
.23 (given that we select a promoted
individual) Probability of probability of being
promoted if you are male 50/120
.417 Probability of probability of being promoted
if you are female 15/80 .188 Overall
probability of being promoted 65/200 .325
Note that the three conditional probabilities
surrounding promotional status and sex are not
the same Therefore, promotion and sex are not
statistically independent
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Probability Distributions
First - two new types of probability
Discrete - have discrete random variables - can
determine the exact number of variable - eg.
Number of animals/plants, Continuous - have an
infinite number of values -e.g. temperature,
weight
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Probability Distributions
First - two new types of probability
Discrete - have discrete random variables - can
determine the exact number of variable - eg.
Number of animals/plants, Continuous - have an
infinite number of values -e.g. temperature,
weight
Each of these kinds of variables will have
an associated probability model
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First Probablility Model - Binomial Model-

• Conditions to be met
• Need to have discrete random variables
• Need to have a Bernoulli process

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A Bernoulli Process
-developed by Jakob Bernoulli

• 3 Conditions
• Must be able to classify outcomes into 2 mutually
  exclusive and exhaustive categories

e.g. for 2 alternatives - boy or girl, head or
tail e.g. for gt2 alternatives - dice - 3 or not
3 - political parties - Liberal or not Liberal
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A Bernoulli Process
-developed by Jakob Bernoulli

• 3 Conditions
• Must be able to classify outcomes into 2 mutually
  exclusive and exhaustive categories
• Trials must be independent

e.g. when flipping a coin - the outcome of the
first flip, say a head, has no effect on the
outcome of the second flip
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A Bernoulli Process
-developed by Jakob Bernoulli

• 3 Conditions
• Must be able to classify outcomes into 2 mutually
  exclusive and exhaustive categories
• Trials must be independent
• Probability of an outcome must remain constant
  from trial to trial

e.g. the chance of getting a head in coin
flipping must be the same on every flip -not
necessarily 0.5 but just the same every time
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A Bernoulli Process
-developed by Jakob Bernoulli

• 3 Conditions
• Must be able to classify outcomes into 2 mutually
  exclusive and exhaustive categories
• Trials must be independent
• Probability of an outcome must remain constant
  from trial to trial

In other words, a Bernoulli trial has outcomes
that are statistically independent
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Example of a non-Bernoulli process Drawing poker
chip from an urn - without replacement
Imagine an urn with five chips - three red and
two blue
Probability of getting red on each of the first
three draws
First draw
3/5
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Example of a non-Bernoulli process Drawing poker
chip from an urn - without replacement
Imagine an urn with five chips - three red and
two blue
Probability of getting red on each of the first
three draws
First draw
3/5
Second draw
2/4
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Example of a non-Bernoulli process Drawing poker
chip from an urn - without replacement
Imagine an urn with five chips - three red and
two blue
Probability of getting red on each of the first
three draws
First draw
3/5
Second draw
2/4
Third draw
1/3
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Example of a non-Bernoulli process Drawing poker
chip from an urn - without replacement
Imagine an urn with five chips - three red and
two blue
Probability of getting red on each of the first
three draws
First draw
3/5
Probability is not the same from trial to trial -
violates the third rule about Bernoulli Processes
Second draw
2/4
Third draw
1/3
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Example of a non-Bernoulli process - a little
more biological
Population of 100 frogs 60 males 40 females
If youre collecting frogs (i.e. catching without
replacement) the probability of catching either
sex will change every time
Trial number 1 Chance of catching a
female 40/100 .4 2 Chance of
catching a female 39/99 .393
3 Chance of catching a male 60/98 .612
4 Chance of catching a female 38/97
.391
Again, the probability changes from trial to trial
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Computing probabilities in a Binomial Model

• Conditions to be met
• Need to have discrete random variables
• Need to have a Bernoulli process

Remember

• If the probabililty of having a son is 0.4 and a
  daughter is 0.6, what is the exact probability of
  having two boys and 1 girl?
• Solution
• Arrange the outcome in any sequence
• B B G
• 2. Compute the probability of obtaining the
  sequence
• P(boy boy girl) .4 x .4 x .6 0.096
• 3. Determine the number of ways the outcome can
  occur - in our case - 3
• B B G, G B B and B G B
• or use the formula
• (Total of outcomes)! (3)! 3
• (Number of event A)! X (Number of event B)!
  (2)! x (1)!
• 4. Multiply the probability of one sequence x the
  number of possible sequences
• 3 x 0.096 .288 or 28.8 of all families

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Using the same procedure, we can calculate
that P(0 boys) .216, P(1 boy) .432, P(2
boys) .288, P(3 boys) .064 Which gives this
distribution
0.5
0 1 2 3
Number of boys
Note that the distribution is skewed compared to
one generated if P(boy) P(girl) 0.5
0.5
0 1 2 3
Number of boys
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Since the Binomial Distribution is a
distribution, it has measures of central tendency
and dispersion. In our boy/girl example, you can
ask - what is the expected number of boys in 3
children (or what is the average number of boys
in three children)? - this is weighted mean like
your lottery example. X (0 x .216) (1 x
.432) (2 x .288) (3 x .064) 1.2
Therefore, the average number of boys in a
family of 3 is 1.2 (with the probability of
having a boy being .4) Variance (s2)
(0-1.2)2(.216) (1-1.2)2(.432)
(2-1.2)2(.288) (3-1.2)2(..064)
.72 Standard deviation (s) v.72
.85
Just for comparison, if the probability of having
a boy is .5, the mean would be 1.5 and the s.d.
.68
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Computing probabilities in a Binomial Model
If the

• If the probabililty of having a son is 0.4 and a
  daughter is 0.6, what is the exact probability of
  having two boys and 1 girl?
• Solution
• Arrange the outcome in any sequence
• B B G
• 2. Compute the probability of obtaining the
  sequence
• P(boy boy girl) .4 x .4 x .6 0.096
• 3. Determine the number of ways the outcome can
  occur - in our case - 3
• B B G, G B B and B G B
• 4. Multiply the probability of one sequence x the
  number of possible sequences
• 3 x 0.096 .288 or 28.8 of all families