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ASU MAT 142 - Basic Rules of Probability

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Basic Rules of ProbabilityThe AxiomsMore AxiomsExamples#61 p.155ANSWERSBasic Rules of ProbabilitySection 3.3The AxiomsIf E is an event and is a subset of S, the sample space, then the following are true:1. 0 ≤ P(E) ≤1 The probability of an event E is between 0 and 1 inclusive.2. P( ) = 0 The probability of an empty set is zero. Consequence: IF P(A ∩ B) = 0 then A ∩ B = which implies A and B are mutually exclusive.3. P(S)=1 The probability of the sample space is 1.More Axioms4. The rule for unions in general is P(A U B) = P(A) + P(B) – P(A ∩ B)5. If A and B are mutually exclusive then P(A U B) = P(A) + P(B).6.The rule for complimentary events is P(E’) = 1 – P(E).Examples•Let E be the event of tossing two dice such that the sum of the face is even. Let F be the event that the sum of the faces is greater than 9. •What is the probability of the sum of the faces being even and greater than 9.•What we are looking for is P(E ∩ F). The only rolls that correspond to that event are (6,4), (4,6), (5,5), (6,6).•Thus P(E ∩ F)=4/36=1/9.•What is the probability of the sum being even or greater than 9.•Want to compute P(E U F).•USE AN AXIOM P(E U F)=P(E)+P(F)-P(E ∩ F)=1/2+1/6-1/9=5/9#61 p.155•Of all flashlights in a large shipment, 15% have a defective bulb, 10% have a defective battery, and 5% have both defects. If you purchase a flashlight from the shipment what are the probabilities of the following:•A defective light bulb or a defective battery,•A good bulb or a good battery,•A good bulb and a good battery.ANSWERS•L be the event of having a bad light bulb.•B be the vent of having a bad battery.•L’ is the event of having a good light bulb.•B’ is the event of having a good battery.•P(L)=0.15, P(B)=0.10, P(L ∩ B)=0.05•The probability of having a bad bulb or bad battery is P(L U B)=P(L)+P(B)- P(L ∩ B)= 0.15+0.10-0.05=0.20=20%.•The probability of having a good bulb or good battery is P(L’ U B’).•Using de’Morgan’s law L’ U B’=(L ∩ B)’•Thus P(L’ U B’) = P((L ∩ B)’)=1-P(L ∩ B)=1-0.05 = 0.95 = 95%•The last part where we have a good bulb and good battery is P(L’∩B’)=P((L U B)’)=1-P(L U


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ASU MAT 142 - Basic Rules of Probability

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