special lecture: conditional probability
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Special Lecture: Conditional Probability. Example of Conditional Probability in the real world: This chart is from a report from the CA Dept of Forestry and Fire Prevention. It shows the probability of a structure being lost in a forest fire given its location in El Dorado - PowerPoint PPT PresentationTRANSCRIPT
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Special Lecture: Conditional Probability
Don’t forget to sign in for credit!
Example of Conditional Probability in the real world:
This chart is from a report from the CA Dept of Forestry and Fire Prevention.It shows the probabilityof a structure beinglost in a forest fire given itslocation in El Dorado county. (calculated usingfuel available, land slope, trees, neighborhood etc.)
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The Plan…Today, I plan to cover material
related to these ALEKS topics.
Specifically, we’ll…• Review all the formulas we’ll
need.• Go over one conceptual
example in depth.• Work through a number of the
ALEKS problems that have been giving you trouble.
• Address any specific questions/problems.
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Formulas:Event Probability Terms/Explanation
A p(A) [0,1] probability of A is between 0 and 1
Not A p(A’) = 1 - p (A) Compliment: Note that the probability of either getting A or not getting A sums to 1.
A or B (or both)
p(AB) = p(A) + p(B)-p(AB)=p(A) + p(B)
Union:
if A & B are mutually exclusive
A & B p(AB) = p(A)p(B) = p(A|B)p(B)
Intersection: only if A and B are independent
A given B
P(A|B) = p(AB)/p(B) Conditional Probability: The probability of event A given that you already have event B.
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Bayes’ Theorem: This is simply derived from what we already know about conditional probability.
Formulas:
p(A|B) = p(B|A)*p(A) p(B)
Or if we don’t have p(B) we can use the more complicated variation of Bayes’:
p(A|B) = p(B|A)*p(A) p(B|A)*p(A) +p(B|A’)*p(A’)
The reason those two formulas are the same has to do with the Law of Total Probabilities:
For any finite (or countably infinite) random variable,
p(A) = ∑ p(ABn)or, p (A) = ∑ p(A|Bn)p(Bn)
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Formulas: All together nowEvent Probability Terms/Explanation
A p(A) [0,1]p (A) = ∑ p(ABn) = ∑ p(A|Bn)p(Bn)
probability of A is between 0 and 1And is the sum of all partitions of A
Not A p(A’) = 1 - p (A) Compliment: probability of either getting A or not getting A sums to 1.
A or B (or both)
p(AB) = p(A) + p(B)-p(AB)=p(A) + p(B)
Union: only if A & B are mutually exclusive
A & B p(AB) = p(A)p(B) = p(A|B)p(B) = p(B|A)p(A)
Intersection: only if A and B are independent
A given B
P(A|B) = p(AB)/p(B) = p(B|A)p(A)/p(B) = p(B|A)p(A) p(B|A)*p(A) +p(B|A’)*p(A’)
Conditional Probability: The probability of event A given that you already have event B.
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Shapes DemoImagine that we have the following population of shapes:
Notice that there are several dimensions that we could use to sort or group these shapes:
• Shape• Color • Size
We could also calculate the frequency with which each of these groups appears and determine the probability of randomly selecting a shape with a particular dimension from the larger set of shapes.
So let’s do that…
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Shapes Demo
• P(R)• P(Y)• P(B)
= 8/24 = 1/3= 8/24 = 1/3= 8/24 = 1/3
Imagine that we have the following population of shapes:
• P( )• P( )• P( )• P( )
= 6/24 = 1/4= 6/24 = 1/4= 6/24 = 1/4= 6/24 = 1/4
• P(BIG)• P(small)
= 12/24 = 1/2= 12/24 = 1/2
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• P(R)• P(Y)• P(B)
= 8/24 = 1/3= 8/24 = 1/3= 8/24 = 1/3
• P( )• P( )• P( )• P( )
= 6/24 = 1/4= 6/24 = 1/4= 6/24 = 1/4= 6/24 = 1/4
• P(BIG)• P(small)
= 12/24 = 1/2= 12/24 = 1/2
Now that we’ve figured out the probability of these events,What else can we do?
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• P(R)• P(Y)• P(B)
= 8/24 = 1/3= 8/24 = 1/3= 8/24 = 1/3
• P( )• P( )• P( )• P( )
= 6/24 = 1/4= 6/24 = 1/4= 6/24 = 1/4= 6/24 = 1/4
• P(BIG)• P(small)
= 12/24 = 1/2= 12/24 = 1/2
Now that we’ve figured out the probability of these events,What else can we do? Lots of stuff!
What’s the probabilityof getting a blue triangle?
= p(B )
= 8/24 * 6/24 = 48/576 = 2/24 = 1/12
= p(B)*p( ) p( )
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• P(R)• P(Y)• P(B)
= 8/24 = 1/3= 8/24 = 1/3= 8/24 = 1/3
• P( )• P( )• P( )• P( )
= 6/24 = 1/4= 6/24 = 1/4= 6/24 = 1/4= 6/24 = 1/4
• P(BIG)• P(small)
= 12/24 = 1/2= 12/24 = 1/2
Now that we’ve figured out the probability of these events,What else can we do? Lots of stuff!
What else?
= p(B ) = 1/12 p( )
p( or B or ) = p(B )
= p(B )+p( )- p(B ) = 8/24 +6/24 - 1/12 =12/24 =1/2
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• P(R)• P(Y)• P(B)
= 8/24 = 1/3= 8/24 = 1/3= 8/24 = 1/3
• P( )• P( )• P( )• P( )
= 6/24 = 1/4= 6/24 = 1/4= 6/24 = 1/4= 6/24 = 1/4
• P(BIG)• P(small)
= 12/24 = 1/2= 12/24 = 1/2
Now that we’ve figured out the probability of these events,What else can we do? Lots of stuff!
What else?
= p(B ) = 1/12 p( )
p( or B or ) = p(B )=1/2
p( given that we have B)
= p(B ) /p(B) = 2/24 / 8/24 = 2/8 = 1/4
= p( |B)
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So, the calculations work out…
But do they make sense??
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How to approach ALEKS problems
1. Write down everything you know.2. Write down (and probably draw out) what
you need to figure out.3. Figure out a plan.4. Go.
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So, Let’s Try an ALEKS problem.
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Any other questions or concerns?