# Conditional probability. Math 425 Introduction to Probability Lecture 9. Example. Conjunctions

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1 Review: Conditioning Rule and Conjunctions Conditional probability Math 425 Introduction to Probability Lecture 9 Definition (Conditioning Rule) Let E and F be events. The conditional probability of E given F is defined by: Kenneth Harris Department of Mathematics University of Michigan provided P (F) > 0. P(E F) P (E F). P (F ) January 28, 2009 Note. From now on I will assume P (F) > 0 when I write P(E F). Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 9 January 28, / 33 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 9 January 28, / 33 Conjunctions Review: Conditioning Rule and Conjunctions Review: Conditioning Rule and Conjunctions We can often compute P (F) and P(E F) easily. If so, we can compute the conjunction P (E F) as well. Lemma (Multiplication Rule) For any events E and F (where P (F) > 0), P (E F ) P(F E) A drawer contains 5 red socks and 3 blue socks. If you remove the socks at random, what is the probability of holding a blue pair? You could do this by enumerating all the possible combinations in the sample space for this experiment, but conditioning provides a simpler method. Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 9 January 28, / 33 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 9 January 28, / 33

2 Review: Conditioning Rule and Conjunctions continued Partition Rule 5 red and 3 blue socks. Consider the two events B i : the ith sock picked is blue (i 1, 2). We want to determine P (B 1 B 2 ). Use the multiplication rule, P (B 1 B 2 ) P (B 1 ) P(B 2 B 1 ) Reason: If you choose 1 blue sock then there are 2 blue socks and 7 socks remaining. So, P(B 2 B 1 ) 2 7. It is often easier to compute the probability of an event by dividing the sample space into two disjoint groups. (Compare to Ross, equation 3.3.1, p. 72). Theorem (Partition Rule) P(E F ) P (F ) + P(E F c ) P (F c ) Proof. Since E F and E F c are mutually disjoint P (E F) + P (E F c ) P(E F) P (F) + P(E F c ) P (F c ) The last line is by the Conditioning Rule. Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 9 January 28, / 33 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 9 January 28, / 33 1: tests 1 continued In a population of individuals a proportion p are subject to a disease (such as AIDS). A test is available which indicates whether an individual has the disease (a positive result). No test is perfect though. Suppose the following: The probability of positive test when an individual has the disease is 95%, (So 5% of the time the test fails to indicate the disease a false negative.) The probability of positive test when an individual does not have the disease is 5% false positive. What is the probability that a randomly selected individual is positive? Solution. Let P be the event that the result is positive and D the event the person has the disease. Then, Use the Partition Rule: P (D) p P(P D) 0.95 P(P D c ) 0.05 P (P) P(P D) P (D) + P(P D c ) P (D c ) 0.95p (1 p) 0.9p You can expect alot of positive results, most of which will be false positives, if the disease is rare (i.e., p is small). Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 9 January 28, / 33 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 9 January 28, / 33

3 Extended Partition Rule It is often more convenient to divide a sample space into several groups. (Compare to Ross, equation 3.3.4, p. 81). Theorem (Extended Partition Rule) Let E be some event and suppose F 1, F 2,..., F n is a collection of mutually exclusive events, one of which must occur: Then, n F i S. k1 n P(E F k ) P (F k ). k1 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 9 January 28, / 33 Extended Partition Rule Proof. We can break the event E into cases (since some one of the events F k must occur): n E (E F k ). k1 The events E F k are mutually exclusive (since the events F k are). By the Sum Rule n P (E F k ) k1 n P(E F k ) P (F k ) k1 The last line is by the Multiplication Rule. Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 9 January 28, / 33 2: coins 2 continued Solution. Let D, T, N be the event of choosing a double-headed, double-tailed and normal coin. Let H be the event that the coin shows heads. We are given the following You have 3 double-headed coins, 1 double-tailed coin and 5 normal coins. You select one coin at random and flip it. What is the probability of heads? So, P (D) 1 3 P (T ) 1 9 P (N) 5 9 P(H D) 1 P(H T ) 0 P(H N) 1 2. P (H) P(H D) P (D) + P(H T ) P (T ) + P(H N) P (N) Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 9 January 28, / 33 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 9 January 28, / 33

4 Converting conditional probabilities The following rule is the of. Theorem (Rule for Converting Conditional Probabilities) Let E and H be events. Then P(H E) Proof. Apply the Conditioning Rule twice is one of the most important in all probability. Theorem () P(H E) Proof. By the Partitioning Rule + P(E H c ) P (H c ) + P(E H c ) P (H c ) P(H E) P (E H). Plug this into the Rule for Converting Conditional Probabilities P(H E) Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 9 January 28, / 33 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 9 January 28, / 33 1 false positives 1 continued In a population of individuals a proportion p are subject to a disease: The probability of positive test when an individual has the disease is 95%, (The cases where the test fails to indicate the disease is called a false negative.) The probability of positive test when an individual does not have the disease is 5%. (This is called a false positive). Should we implement universal (or random) testing? That is, should we be confident a positive test result indicates the disease when we test everyone? Solution. Let D be the event of having a disease and R be the event of a positive result. We are given P (D) p P(R D) 0.95 P(R D c ) 0.05 P (R) 0.9p (We computed the last in 1 of the previous section.) You want to know P(D R). Use P(D R) P(R D)P (D) P (R) 0.95p 0.9p Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 9 January 28, / 33 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 9 January 28, / 33

5 1 continued Extended Bayes Rule D: disease, R: positive test result P(D R) 0.95p 0.9p If the disease is rare: p (about the rate of AIDS in the US). Then P(D R) % of positive cases will be false positives. If the disease is common: p 0.1. Then P(D R) 0.69 Two-thirds of positives are now people with the disease. The extended version of Bayes Theorem just applies the Extended Partition Rule to the Rule for Converting Conditional Probabilities. (See Ross Proposition 3.1, p. 81.) Theorem (Extended ) Let E be some event and suppose H 1, H 2,..., H n is a collection of mutually exclusive events, one of which must occur: Then, P(H i E) n H k S. k1 P(E H i ) P (H i ) n k1 P(E H k) P (H k ). Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 9 January 28, / 33 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 9 January 28, / 33 3: An urn problem An urn contains two balls, which have been randomly chosen to be either red or blue. We perform the following experiment to determine the composition of the urn. 1 Select a ball and record its color. 2 Replace the ball in the urn. 3 Mix the contents of the urn well. Suppose we perform this experiment twice, and each time get a red ball. What is the most likely composition of the urn? 3 continued Solution. Prior to performing the experiment we have three hypotheses about the urn: D: the balls in the urn have different colors, R: both balls in the urn are red, B: both balls in the urn are blue. Since the balls were randomly chosen to be placed in the urn, we have the following probabilities (prior to conducting the experiment): P (D) 1 2 P (R) P (B) 1 4. Our experiment produced the following outcome: R 2 : Two red balls are drawn. The likelihood of this event given each hypothesis P(R 2 D) 1 4 P(R 2 R) 1 P(R 2 B) 0 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 9 January 28, / 33 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 9 January 28, / 33

6 3 continued Use the Partition Rule: P (R 2 ) P(R 2 D)P (D) + P(R 2 R)P (R) + P(R 2 B)P (B) Use to recompute the probability of each hypothesis: P(D R 2 ) P(R 2 D) P (D) P (R 2 ) P(R R 2 ) P(R 2 R) P (R) P (R 2 ) P(B R 2 ) P(R 2 B) P (B) P (R 2 ) It is useful to know that conditional probabilities obey the probability axioms. (See Ross, Proposition 3.5.1, p. 102) Theorem Let F be any event with P (F) > 0. Then, the function P( F ) on the event space S is a probability function. That is, P( F) satisfies the probability axioms. 1 0 P(E F) 1 for all events E, 2 P(S F) 1, 3 If E 1 and E 2 are mutually exclusive events, then P(E 1 E 2 F) P(E 1 F) + P(E 2 F) The most likely explanation is that the balls in the urn are both red. Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 9 January 28, / 33 Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 9 January 28, / 33 Proof : Cesarian sections For any event F with P (F) > 0: ➀ Since E F F, ➁ 0 P(S F) P (E F) P (F) P (S F) P (F ) P(E F) 1. P (F) P (F) 1 ➂ If E 1, E 2 are mutually exclusive, then so is E 1 F and E 2 F. By the distributive law So, (E 1 E 2 ) F (E 1 F ) (E 2 F) P(E 1 E 2 F) P (E 1 F) + P (E 2 F ) P (F ) P(E 1 F) + P(E 1 F ) Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 9 January 28, / 33 98% of all babies survive delivery. However, 15% of all deliveries involve Cesarian (C) sections, and then the survival rate drops to 96%. If a randomly chosen pregnant woman does not have a C section, then what is the probability that the baby survives? Let B be the event that the baby survives and C the event that a C section is performed. We want to compute P(B C c ) Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 9 January 28, / 33

7 continued We are given the following data P (B) 0.98 P (C) 0.15 P(B C) By converting conditional probabilities P(B C c ) P(Cc B) P (B) P (C c ) The right-side can be computed by converting again P(C B) P(B C) P (C) P (B) P(C c B) 1 P(C B) The baby s probability of surviving when no C section is performed: P(B C c ) Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 9 January 28, / 33 : prostate cancer Prostate cancer is a common type of cancer found in men. One test for prostate cancer measures the level of a protein PSA (prostate specific antigen) produced only by the prostate gland. The test is notoriously unreliable though. The probability that a noncancerous man will have elevated PSA is 13.5%, with this probability increasing to 26.8% if the man does have cancer. Suppose a doctor believes that a certain patient has probability p of having prostrate cancer, before testing PSA levels. (a) What is the probability of his having cancer if he has elevated PSA levels? (b) What is the probability of his having cancer if he does not have elevated PSA levels? Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 9 January 28, / 33 continued Let E be the event of having an elevated PSA and H be the hypothesis the patient has cancer. We are given the following P (H) p P(E H) P(E H c ) We want to compute (a) P(H E) and (b) P(H E c ). Use to compute: (a) : P(H E) (b) : P(H E c ) P(E c H) P (H) P (E c ) We need only compute. Here we use the Partition Rule: + P(E H c ) P (H c ) p (1 p) p P (E c ) p Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 9 January 28, / 33 continued p P (E c ) p (a) : P(H E) (b) : P(H E c ) P(E c H) P (H) P (E c ) 0.732p p 0.268p p (1 P(E H)) P (H) P (E c ) Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 9 January 28, / 33

8 completed P(H E) 0.268p p P(H E c ) 0.732p p If the doctor is confident the patient has cancer: p P(H E) P(H E c ) If the doctor is riding the fence: p 0.5. P(H E) P(H E c ) Testing PSA levels is not definitive on its own. Kenneth Harris (Math 425) Math 425 Introduction to Probability Lecture 9 January 28, / 33

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