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Unformatted text preview: ignition are stolen. On the other hand, only 0.2% of the cars without keys left in the ignition are stolen. Suppose a car in Anytownville is stolen. What is the probability that the keys were left in the ignition? P(Keys) = 0.10, P(Keys ' ) = 1 0.10 = 0.90. P(Stolen Keys) = 0.042. P(Stolen Keys ' ) = 0.002. Need P(Keys Stolen) = ? Bayes’ Theorem: P(Keys Stolen) = ) Keys P(Stolen ) P(Keys Keys) P(Stolen P(Keys) Keys) P(Stolen P(Keys)    ' ' × + × × = 002 . 90 . 042 . 10 . 042 . 10 . × + × × = 0.70 . OR Stolen ' Stolen 0.042 ⋅ 0.10 Keys 0.0042 0.0958 0.10 Keys ' 0.002 ⋅ 0.90 0.0018 0.8982 0.90 0.0060 0.9940 1.00 P(Keys Stolen) = 0060 . 0042 . P(Stolen) Stolen) P(Keys = ∩ = 0.70 . OR P(Stolen) = 0.0042 + 0.0018 = 0.0060. P(Keys Stolen) = 0060 . 0042 . P(Stolen) Stolen) P(Keys = ∩ = 0.70 ....
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 Spring '08
 Kim
 Statistics, Conditional Probability, Probability, Automobile, Bayesian probability, Anytownville

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