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.)

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.

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.

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)

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.

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…

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

• 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?

• 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( )

• 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

• 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)

So, the calculations work out…

But do they make sense??

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.

So, Let’s Try an ALEKS problem.

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Any other questions or concerns?