Special Lecture: Conditional ProbabilityThe Plan…Formulas:Slide 4Formulas: All together nowShapes DemoSlide 7PowerPoint PresentationSlide 9Slide 10Slide 11So, the calculations work out…How to approach ALEKS problemsSo, Let’s Try an ALEKS problem.Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Any other questions or concerns?Special Lecture: Conditional ProbabilityDon’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/ExplanationAp(A)  [0,1]probability of A is between 0 and 1Not Ap(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(AB) = p(A) + p(B)-p(AB)=p(A) + p(B)Union: if A & B are mutually exclusiveA & Bp(AB) = p(A)p(B) = p(A|B)p(B)Intersection: only if A and B are independent A given BP(A|B) = p(AB)/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(ABn)or, p (A) = ∑ p(A|Bn)p(Bn)Formulas: All together nowEvent Probability Terms/ExplanationAp(A)  [0,1]p (A) = ∑ p(ABn) = ∑ p(A|Bn)p(Bn)probability of A is between 0 and 1And is the sum of all partitions of ANot Ap(A’) = 1 - p (A)Compliment: probability of either getting A or not getting A sums to 1.A or B (or both)p(AB) = p(A) + p(B)-p(AB)=p(A) + p(B)Union: only if A & B are mutually exclusiveA & Bp(AB) = p(A)p(B) = p(A|B)p(B) = p(B|A)p(A)Intersection: only if A and B are independent A given BP(A|B) = p(AB)/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 = 1/3Imagine 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 = 1/2Now 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 = 1/2Now 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 = 1/2Now 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 = 1/2Now 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 problems1. 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.√√√√√√√√√√√√√√√√√√√√√Any other questions or