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1 Probability Theory Dr. Deshi Ye yedeshi@zju.edu.cn

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2 Outline: 1. Conditional Probability 2. Bayes’ Theorem

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3 3.6 Conditional probability Probability of an event is meaningful iff it refers to a given sample. P(A|S): the probability of A given some space S. If respect to more samples.

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4 D and I 10 Ex. 500 machines. Improper assemble I= 30;Defective D= 15; Both I and D = 10. P(D)=? P(D|I)=? P(D and I)=? I 20 D and I 10

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5 Conditional prob. Theorem. If A and B are any events is S and P(B) is not empty, the conditional probability of A given B is:

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6 Ex. A coin is flipped twice. If we assume that all four points in the sample space are equally likely, what is the probability that both flips result in heads, given that the first flip does? Solution: Let E = {(H,H)} be the event that both flips land heads, and F={(H,H), (H,T)} denote the event that the first flip lands heads, then the desired probability is given by

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7 Ex. Suppose a family has two children. We assume that the probability of having a baby boy is ½. Now suppose we wish to find the probability that the family has one boy and one girl, but we also have the information that at least one of the children is a boy. What is the probability?

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8 Ex. Celine is undecided as to whether to take a French course or a chemistry course. She estimates that her probability of receiving an A grade would be ½ in a French course, and 2/3 in a chemistry course. If Celine decides to base her decision on the flip of a fair coin, what is the probability that she gets an A in chemistry?

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9 Solution If we let C be the event that Celine takes chemistry and B denote the event that she receive an A in whatever course she takes, then the desired probability is P(CB).

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10 Multiplication rule

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11 Multiplication rule – Ex. An ordinary deck of 52 playing cards is randomly divided into 4 piles of 13 cards each. Compute the probability that each pile has exactly 1 Ace?

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12 Solution: another approach Define events E1 = {the ace of spades is in any one of the piles} E2 = {the ace of spades and the ace of hearts are in different piles} E3 = {the aces of spades, hearts, and diamonds are all in different piles} E4 = {all aces are in different piles} P(E1) = 1, P(E2|E1)=39/51, P(E3|E1E2)=26/50, P(E4|E1E2E3) = 13/49. P(E1E2E3E4) = P(E1)P(E2|E1)P(E3|E1E2)P(E4|E1E2E3) = 0.105

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13 Independent A is independent of B if and only if P(A|B)=P(A) or P(A and B) = P(A) * P(B)

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14 EX. Suppose that we toss 2 dice. Let E denote the event that the sum of the dice is 6 and F denote the event that the first die equals 4. Q: are the events E and F independent?

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15 Solution PP(EF) = P({4, 2}) = 1/36. PP(E) = 5/36 P(F) = 1/6 PP(E)P(F) = 5/216 CConclusion: E and F are not independent!! QQuestion: how about the event E changed to: the sum is 7.

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16 Properties IIf E and F are independent, then so are E and IIf E is independent of F and is also independent of G. Is E then necessarily independent of FG? AAns: No!!

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17 Generalization Theorem 3.8 If A and B are any events in S, then if P(A)>0,P(B)>0. If independent

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18 Discussion P(A|B) and P(A ∩ B) A, B are independent, mutually exclusive, what is the difference?

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19 Bayes’ Theorem Event A

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20 Bayes’ Theorem An experiment depends on the outcomes of various intermediate stages. EX. An assembly plant receives its voltage regulators. 60% from B1, 30% from B2, 10% from B3 Perform according to specifications: B1: 95%, B2: 80%, B3: 65%. Event: a voltage regulator received by the plant performs according to specifications.

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21 Tree diagram 0.6 0.3 0.1 B1 B2 B3 P(A|B1)=0.95 A A A P(A|B2)=0. 8 P(A|B3)=0.65

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22 Solution

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23 Rule of total probability Theorem 3.10. If are mutually exclusive events of which one must occur, then

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24 Special case

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25 EX. A laboratory blood test is 95 percent effective in detecting a certain disease when it is, in fact, present. However, the test also yield a “false positive” result for 1 percent of the health person tested. Q: If.5 percent of the population actually has the disease, what is probability a person has the disease given that the test result is positive?

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26 Solution Let D be the event that the tested person has the disease. Let E be the event that the test result is positive. The desired probability is P(D|E) Only 0.323!

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27 Bayes’ Theorem Theorem. If are mutually exclusive events of which one must occur, then “Effect” A was “caused” by the event B r The prob. Of reaching A via the r-th branch of the tree

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28 EX.

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29 Expectation Expectation: If the probability of obtaining the amounts then the mathematical expectation is

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30 Motivations The expected value of x is a weighted average of possible values that X can take on, each value being weighted by the probability that X assumes it. Frequency interpretation

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31 Summary Conditional probability Independent of events A, B Bayes’ rule, B i disjoint

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Unit 4 Section 3.1.

Unit 4 Section 3.1.

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