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For Thursday, Feb. 20 Bring Conditional Probability Worksheet to class Exam #1:March 4 Project Presentations Start March 6 Team Homework #4 due Feb. 25 READ: Project 1: Baye’s Theorem FOCUS LESSON

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Conditional Probability P(A) represents the probability assigned to A -- it is the “unconditional” probability Sometimes there may be conditions that affect the probability assigned to A; “an event B has occurred” The conditional probability of an event, A, given that an event B has happened is denoted P(A | B) P(A | B) is read “the probability that A occurs given that B has occurred”

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Conditional Probability, con’t Given that B has occurred, the relevant sample space has changed; it is no longer S but consists only of the outcomes in B For any events A and B with P(B)>0,

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Conditional Probability, con’t

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We can solve these problems using several methods: Use the formula Venn Diagrams Frequency Tables Tree Diagrams

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Example-Using definition The probability that event A occurs is.63. The probability that event B occurs is.45. The probability that both events occur is.10. Find by using the definition: P(A | B) P(B | A)

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Exercise #1 Suppose that A and B are events with probabilities: P(A)=1/3, P(B)=1/4, P(A ∩ B)=1/10 Find each of the following using a Venn Diagram: 1.P(A | B) 2.P(B | A) 3.P(A C | B) 4.P(B C | A) 5.P(A C | B C )

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Example Consider the experiment of rolling a fair die twice All outcomes in S are equally likely

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Example, Using Table Let E=the sum of the faces is even Let S 2 =the second die is a 2 Find 1. P(S 2 | E) 2. P(E | S 2 )

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Example, Using Table One way of doing this is to construct a table of frequencies: Event AEvent A c TOTALS Event BTotal B Event B c Total B c Total ATotal A c Grand Total

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Example, con’t Let’s try it to find P(S 2 |E) and P(E|S 2 ) Event EEvent E c TOTALS Event S 2 Event S 2 c Remember:

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Example Let S be the event that a person is a smoker and let D be the even that a person has a disease. The probability that a person has a disease is.47 and the probability that a person is a smoker is.64. The probability that a smoker has a disease is.39. –Find the probability that a person with a disease is a smoker

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Example-Tree Diagram Three manufacturing plants A, B, and C supply 20%, 30% and 50%, respectively, of all shock absorbers used by a certain automobile manufacturer. Records show that the percentage of defective items produced by A, B and C is 3%, 2% and 1%, respectively. –What is the probability that the part came from manufacturer A, given that the part was defective? –What is the probability that the part came from B, given that the part was not defective?

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Independent Events If two events are independent, the occurrence of one event has no effect on the probability of the other. E and F are independent events if P(E ∩ F)=P(E) * P(F) Similarly, P(E ∩ F ∩ G)=P(E)* P(F) * P(G), etc Independence of E and F implies that P(E | F)=P(E) and P(F)= P(F | E) If the events are not independent, then they are dependent.

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Independent Events Consider flipping a coin and recording the outcome each time. Are these events independent? Let H n =the event that a head comes up on the nth toss What is the P(H 1 | H 2 )? What is the probability P(H 1 ∩ H 2 ∩ H 3 )?

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Independent Events You throw two fair die, one is green and the other is red, and observe the outcomes. Let A be the event that their sum is 7 Let B be the event that the red die shows an even # Are these events independent? Explain. Are these events mutually exclusive? Explain

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Independent Events, con’t You throw two fair die, one green and one red and observe the numbers. Decide which pairs of events, A and B, are independent: 1.A: the sum is 5 B: the red die shows a 2 2.A: the sum is 5 B: the sum is less than 4 3. A: the sum is even B: the red die is even

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Conditional Probability and Independence If E, F and G are three events, then E and F are independent, given that G has happened, if P(E ∩ F | G)=P(E | G) *P(F | G) Likewise, events E, F and G are independent, given that H has happened, given that G has happened, if P(E ∩ F ∩ G | H)=P(E | H) * P(F | H) * P(G | H)

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Independence In the manufacture of light bulbs, filaments, glass casings and bases are manufactured separately and then assembled into the final product. Past records show: 2% of filaments are defective, 3% of casings are defective and 1% of bases are defective. What is the probability that one bulb chosen at random will have no defects?

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Focus on The Project So, we have found the probability of success and the probability of failure, based on all the records Using information about our borrower might change our probability of success and failure

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Focus on the Project Let S and F be the events of success and failure, respectively Let Y be the event of having 7 years of experience Let T be the event of having a Bachelor’s Degree Let C be the event of having a normal state of economy

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Focus on the Project In terms of conditional probability, we would like to know P(S | Y), P(S | T), and P(S | C) We would also like to know the corresponding probabilities for failure We can estimate these from our bank records When we find P(S | Y) we are implying we are looking at BR bank, etc.

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Focus on the Project How can we find P(S | Y)? Once we have this value, we can find the other conditional probabilities in the same way With the conditional probabilities, we can find the correlating expected values Based on these expected values, we can revise our decision on whether to foreclose or workout?

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Focus on the Project What potential problem do we encounter when we look at the expected values that we just found? We would like to find P(S | Y T C) and P(F | Y T C) -- unfortunately our bank records do not hold this information so we can’t find it directly So, let’s calculate something that we can find AND will be of importance to us later in finding out the above probabilities

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Focus on the Project Let’s find P(Y T C | S) and then in a similar fashion, P(Y T C | F) The project description says that Y, T, and C are independent events, even when they are conditioned upon S or F. Hence, P(Y T C | S) = P(Y|S)*P(T|S)*P(C|S) P(Y T C | F) is similar

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What do you need to do? Calculate P(Y T C | S) and P(Y T C | F) using your borrower’s information Make sure you are keeping all of your information in an Excel file Once we have these numbers, we are going to learn how to use these numbers to find out what we need to know (but can’t get directly)

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