Note that if A and B are independent then PBAPB Conditional probability

Note that if a and b are independent then pbapb

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Note that if A and B are independent, then P(B|A)=P(B) Conditional probability: 𝑃𝑃 𝐵𝐵 𝐴𝐴 = 𝑃𝑃 ( 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝐵𝐵 ) 𝑃𝑃 ( 𝐴𝐴 )
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Conditional Probability and the Prosecutor Fallacy P(guilty and fits description) = P(fits description| guilty) * p(guilty) = P(fits description | guilty) = 1 P(guilty) = 1/ 10.000.000 P(fits description) = 0.000002 Hence: P(guilty | fit description) = 1∗ ( 1 / 10 . 000 . 000 ) 0 . 00002 = 1/20
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So, this is a nice theoretical example? NO! The Lucia de Berk case in the Netherlands (2003) Paediatric Nurse Life imprisonment for 7 murders and 3 attempted nurse Exonerated in 2010 The Sally Clark case in the UK (1999) Mother of 2 sons Life imprisonment for murdering her sons Sudden death syndrome Exonerated in 2003 Further watching:
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Bayes’ rule Suppose you know P(A), P(B|A) and 𝑃𝑃 𝐵𝐵 𝐴𝐴 𝑐𝑐 and you want to calculate P(A|B) (see textbook, example 5.10). Then the definition of conditional probability gives: P A B = 𝑃𝑃 𝐵𝐵 𝐴𝐴 ∗𝑃𝑃 ( 𝐴𝐴 ) 𝑃𝑃 ( 𝐵𝐵 ) And 𝑃𝑃 𝐵𝐵 = 𝑃𝑃 𝐵𝐵 𝐴𝐴 ∗ 𝑃𝑃 𝐴𝐴 + 𝑃𝑃 𝐵𝐵 𝐴𝐴 𝑐𝑐 ∗ 𝑃𝑃 𝐴𝐴 𝑐𝑐 Inserting P(B) in the denominator gives us Bayes’ rule: 𝑃𝑃 𝐵𝐵 𝐴𝐴 ∗ 𝑃𝑃 ( 𝐴𝐴 ) 𝑃𝑃 𝐴𝐴 ∗ 𝑃𝑃 𝐵𝐵 𝐴𝐴 + 𝑃𝑃 𝐴𝐴 𝑐𝑐 ∗ 𝑃𝑃 𝐵𝐵 𝐴𝐴 𝑐𝑐
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Let’s play a game! Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the other two a goat.
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Win a car or a goat Let us actually play this game! - illusions.com/simulator/montysim.htm
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The Monty Hall Problem (2) You pick a door, say No.1, and the host (His name is Monty Hall) , who knows what's behind the doors, opens another door, say No.3, which has a goat. Monty then asks: "Do you want to pick door No.2?" Is it to your advantage to switch your choice?
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The Monty Hall Problem (3) It is always the best strategy to switch doors! The probability of winning the car without change of doors is 1/3 The probability of winning the car with change of cars is 2/3. To some people the outcome may be counter-intuitive. To those I have two suggestions: 1. Rephrase the problem such that there are 100 doors and the quiz master opens up 98 after you have chosen a door. Would you change doors then? 2. Let us apply the rules of probability to show that you always have to change doors when you are given the option to change!
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The Monty Hall Problem (4) To show that you have to switch doors, suppose that: you pick door #1, call it: “a”. Monty Hall opens door #2, call it “b”. There is one door left (#3): call it “c”. Let A, B, and C be the events that the car is behind door a, b, or c, respectively.
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