Dominique Mwita

Lecture 5 Conditional Probability, Bayes Theorem, and Independence

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Conditional Probability • (Def ) Probability of event A given that the event B occurred = Conditional probability of A given B

P A∩B
= P A B = P B • Graphical Representation: • Example 1: Suppose that of all individuals buying a certain camera, 60% include an optional memory card in their purchase, 40% include an extra battery, and 30% include both a card and battery. Consider randomly selecting a buyer and let A=memory card purchased and B=battery purchased.

Then, P (A) = P Optional Memory = 0.6; P (B) = P Extra Bat. =
0.4; P (A ∩ B) = P Both optional memory card and extra Battery = 0.3. Then, what is the probability that if one choose an extra bat., he also choose optional memory? What is the probability that if one choose an optional mem., he also choose extra bat.?

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Bayes Theorem
• Multiplicative rule for P A ∩ B




P A ∩ B = P A · P B A = P B · P A B Example 2 A chain of video stores sells three different brands of VCRs. Of its VCR sales, 50% are brand 1(the least expensive), 30% are brand 2, and 1

20% are brand 3. Each manufacturer offers a 1-year warranty on parts and labor. It is known that 25% of brand 1’s VCRs require warranty repair work, whereas the corresponding percentages for brands 2 and 3 are 20% and 10%, respectively. What is the probability that a randomly selected purchaser has bought a brand 1 VCR that will need repair while under warranty? • Law of total probability Let A1, A2, · · · , Ak are mutually exclusive and exhaustive events = a partition of the sample space. Then,



P B = P B ∩ A1 + P B ∩ A2 + · · · + P B ∩ Ak





= P A1 P B A1 + P A2 P B A2 + · · · + P Ak P B Ak =

k X i=1



P Ai P B Ai

Example 2 continued. In example 2, what is the probability that a randomly selected purchaser has a VCR that will need repair while under warranty? • Bayes Theorem
Let A1 , A2, · · · , Ak are a partition of the sample space with P Ai > 0. Then,

P A ∩ B j
P Aj B = P B

P B Aj P Aj = Pk

P Ai · P B Ai i=1

– Example 2 continued. In example 2, if a customer returns to the store with a VCR that needs warranty repair work, what is the probability that it is a brand 1 VCR? A brand 2 VCR? A brand 3 VCR? 2

– Example 3 Only 1 in 1000 adults is afflicted with a rare disease for which a diagnostic test has been developed. The test is such that when an individual actually has the disease, a positive result will occur 99% of the time, whereas an individual without the disease will show a positive test result only 2% of the time. If a randomly selected individual is tested and the result is positive, what is the probability that the individual has the disease?

– Problem 1 At a certain gas station, 40% of the customers use regular unleaded gas(A1), 35% use extra unleaded gas(A2), and 25% use premium unleaded gas(A3). Of those customers using regular gas, only 30% fill their tanks (event B). Of those customers using extra gas, 60% fill their tanks, whereas of those using prenimum, 50% fill their tanks. 1. What is the probability that the next customer will request extra unleaded gas and fill the tank (that is, A2 ∩ B)? 2. What is the probability that the next customer fills the tank? 3. If the next customer fills the tank, what is the probability that regular gas is required? Extra gas? premium gas?

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Independence • (Def ) A and B are independent:

or P A B = P A




P A∩B =P A P B

• Example 4 Five independent components are connected in series. The probability that each component works is 0.9. What is the probability that the system does not work? 3

(sol) Denote a component that fails by F and one that doesn’t fail by S (for success). For A to occur, at least one of the individual components must fail. So, A=SSSSF, SSSFS,. . . which are 31 different outcomes in A. But, consider Ac (or A’), the event that the system works. The event Ac has only one outcome(SSSSS). P (Ac ) = P (SSSSS) = P (S) × P (S) × P (S) × P (S) × P (S) = 0.95 = 0.59. So, P (Ac ) = 1 − P (A) = 1 − 0.59 = 0.41.

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