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Holt Mcdougal Larson Pre-algebra: Student Edition 2012
Holt Mcdougal Larson Pre-algebra: Student Edition 2012
1st Edition
ISBN:9780547587776
Author:HOLT MCDOUGAL
Publisher:HOLT MCDOUGAL
Chapter11: Data Analysis And Probability
Section11.8: Probabilities Of Disjoint And Overlapping Events
Problem 2C
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An HDMI cable in a lost and found bin is in working condition with probability 0.9, whereas a VGA cable found in the same bin is in working condition with probability 0.5. The bin currently contains 6 HDMI cables and 4 VGA cables. Two cables are randomly chosen from the bin at once. (a) What is the probability that both cables work? (b) Given that both cables work, what is the conditional probability that both are HDMI cables
Transcribed Image Text:An HDMI cable in a lost and found bin is in working condition with probability 0.9, whereas a VGA cable found in the same bin is in working condition with probability 0.5. The bin currently contains 6 HDMI cables and 4 VGA cables. Two cables are randomly chosen from the bin at once. (a) What is the probability that both cables work? (b) Given that both cables work, what is the conditional probability that both are HDMI cables
Axioms of Probability Also Note: 1. P(S)=1 For any two events A and B, 2. For any event E, 0< P(E) < 1 P(A) = P(AN B) + P(AN B') 3. For any two mutually exclusive events, and P(EUF) = P(E)+ P(F) P(AN B) P(A|B)P(B). Events A and B are independent if: Addition Rule P(EUF) = P(E)+ P(F) – P(EnF) P(A|B) = P(A) or Conditional Probability P(AN B) = P(A)P(B). Р(BJA) — P(ANB) P(A) Bayes' Theorem: Total Probability Rule Р(A В)P(В) Р(А|B)Р(В) + Р(A|B')P(В') P(A) = P(A|B)P(B)+P(A|B')P(B') P(B|A) Similarly, Similarly, P(A) =P(A|E1)P(E1) + P(A|E2)P(E2)+ ...+ P(A|Ek)P(Ek) P(B|E1)P(E1) P(B|E1)P(E1) + P(B|E2)P(E2) + · .+ P(B|ER)P(Ex) P(E1|B)
Transcribed Image Text:Axioms of Probability Also Note: 1. P(S)=1 For any two events A and B, 2. For any event E, 0< P(E) < 1 P(A) = P(AN B) + P(AN B') 3. For any two mutually exclusive events, and P(EUF) = P(E)+ P(F) P(AN B) P(A|B)P(B). Events A and B are independent if: Addition Rule P(EUF) = P(E)+ P(F) – P(EnF) P(A|B) = P(A) or Conditional Probability P(AN B) = P(A)P(B). Р(BJA) — P(ANB) P(A) Bayes' Theorem: Total Probability Rule Р(A В)P(В) Р(А|B)Р(В) + Р(A|B')P(В') P(A) = P(A|B)P(B)+P(A|B')P(B') P(B|A) Similarly, Similarly, P(A) =P(A|E1)P(E1) + P(A|E2)P(E2)+ ...+ P(A|Ek)P(Ek) P(B|E1)P(E1) P(B|E1)P(E1) + P(B|E2)P(E2) + · .+ P(B|ER)P(Ex) P(E1|B)
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