Chapter 2 - Lecture 4 September 21st, 2009 Chapter 2 - Lecture 4
Chapter 2 - Lecture 4
Probability Until now we have learn how to assign probabilities on sets that are consisted of simple events without assuming that we know anything regarding the environment that might affect the probability. Chapter 2 - Lecture 4
is the probability than an event will occur when we know some facts that have already happen. Example: I have a class of 100 students What is the probability that student A has failed the class? What is the probability that student A has failed the class given that 15 students have failed the class? What is the probability that student A has failed the class given that all the students have failed the class? Chapter 2 - Lecture 4
Outline of event A given that the event B has occurred: P(A B) Chapter 2 - Lecture 4
Example Let say that I have 100 CDs from each of the two brands CDMaker and MakingCD Let say that 5 of the CDs from CDMaker are defective and 10 of the CDs from MakingCD are defective. Denote with A the event that a CD is defective and with B the event that a CD is branded from CDMaker. Find P(A), P(B), P(A c ), P(A B), P(A c B), P(A B c ), P(B A), P(B c A c ). Chapter 2 - Lecture 4
Formula Outline In the previous example we constructed a table to find Conditional probabilities. If we are not able to construct a Table then we can use the following formula to find conditional probabilities: P(A B) = P(A B) P(B) Chapter 2 - Lecture 4
Example Let say 50% of the CDs I own are CDMaker branded and 7.5% are defective and 2.5% are both branded CDMaker and defective. Denote with A the event that a CD is defective and with B the event that a CD is branded from CDMaker. What is the P(the CD is defective the CD is branded CDMaker) = P(A B)? Chapter 2 - Lecture 4
Formula Outline The previous formula gave us another way of calculating the Probability of events A and B, as follows: P(A B) = P(A B)P(B) Chapter 2 - Lecture 4
Example 2.29 page 77 Outline Chapter 2 - Lecture 4
Example 2.29 page 77 (using tree diagrams) Chapter 2 - Lecture 4
Let A 1,..., A k be mutually exclusive and exhaustive events. Then for any event B, P(B) = k P(B A i )P(A i ) i=1 The events A 1,..., A k are said to be exhaustive if one of them must occur, that is A 1... A k = S Chapter 2 - Lecture 4
Let A 1,..., A k be mutually exclusive and exhaustive events with P(A i ) > 0, i Then for any event B for which P(B) > 0 P(A j B) = P(A j B) P(B) = P(B A j)p(a j ) k P(B A i )P(A i ) i=1 Chapter 2 - Lecture 4
Example Denote with A the event that a CD is defective and with B the event that a CD is branded from CDMaker. We know that 7.5% of the CDs I own are defective If a CD is defective there is 33.33% probability that it is branded CDMaker. If it is not defective there is 51.35% that it was made by CDMaker Find P(A B) and P(A c B) first by constructing a tree diagram and then by using the formula. Chapter 2 - Lecture 4
Section 2.4 page 80 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65 Chapter 2 - Lecture 4