16.322 Stochastic Estimation and Control Professor Vander Velde 1. P(ABCD.=P(A)P(B A)P(C|AB)P(D 1 ABC) Derive this by letting A=CD. Then P(BCD)= P(CD)P(B ICD)= P(C)P(DIC)P(DICD) 2. If A,, A2r.. is a set of mutually exclusive and collectively exhaustive events, then P(E)=P(EA)+P(EA)+…+P(EA1)=P(A)P(E|A)+P(A2)P(E|A)+…+P(An)P(E|A) A A3 A P(A+B)=P(A)+P(B)-P(AB)must subtract off P(AB) because it is counted twice by the first two terms of the rhs P(A+B+C)=P(A)+P(B)+P(C)-P(AB)-P(AC)-P(BC)+ P(ABC)(for three events) If four events, P(A+B+C+ D) would be the sum of probabilities of the single events, minus the probability of the joint events taken two at a time, plus the probability of the joint events taken three at a time, minus the probability of the oint event taken four at a time Conclusion: If events are mutually exclusive, express result in terms of the sum of events; if independent, in terms of joint events. Note that this reduces to just the sum of the probabilities of the events if the events are mutually exclusive Example: Simplex system Probability of component failure=P(F) P(system failure)=P(any component fails)=P(F +F2+.+F) Assume component failures are independent. Take advantage of this by working with the complementary event Lecture16.322 Stochastic Estimation and Control Professor Vander Velde Lecture 2 1. P ABCD P A P B A P C AB P D ABC ( ...) ( ) ( | ) ( | ) ( | )... = Derive this by letting A=CD. Then P BCD P CD P B CD P C P D C P D CD ( ) ( )( | ) ()( | )( | ) = = 2. If A1 , A2 ,… is a set of mutually exclusive and collectively exhaustive events, then 1 2 1 12 2 ( ) ( ) ( ) ... ( ) ( ) ( | ) ( ) ( | ) ... ( ) ( | ) P E P EA P EA P EA P A P E A P A P E A P A P E A = + ++ = + ++ n nn P A B P A P B P AB ( ) () () ( ) += + − must subtract off P(AB) because it is counted twice by the first two terms of the RHS. P A B C P A P B P C P AB P AC P BC P ABC ( ) () () () ( ) ( ) ( ) ( ) ++ = + + − − − + (for three events). If four events, PA B C D ( ) +++ would be the sum of probabilities of the single events, minus the probability of the joint events taken two at a time, plus the probability of the joint events taken three at a time, minus the probability of the joint event taken four at a time. Conclusion: If events are mutually exclusive, express result in terms of the sum of events; if independent, in terms of joint events. Note that this reduces to just the sum of the probabilities of the events if the events are mutually exclusive. Example: Simplex system Probability of component failure ( ) = P Fi 1 2 (system failure) (any component fails) ( ... ) P P PF F F = = + ++ n Assume component failures are independent. Take advantage of this by working with the complementary event