MATLAB: An Introduction with Applications
6th Edition
ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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- The prior probabilities for events A₁, A2, and A3 are P(A₁) = 0.20, P(A₂) = 0.30, and P(A3) = 0.50. The conditional probabilities of event B given A₁, A₂, and A3 are P(B | A₁) = 0.50, P(B | A₂) = 0.30, and P(B | A3) = 0.40. (Assume that A₁, A₂, and A3 are mutually exclusive events whose union is the entire sample space.) (a) Compute P(B ʼn A₁), P(B n A₂), and P(B n A3). P(B n A₁) P(B n A₂) = P(B n A3) = P(A₁)P(B | A₁) (b) Apply Bayes' theorem, P(A¡ | B) = 7 to compute the posterior probability P(A₂ | B). (Round your answer to two decimal P(A₁)P(B | A₁) + P(A₂)P(B | A₂) + + P(An)P(B | An)' places.) (c) Use the tabular approach to applying Bayes' theorem to compute P(A₁ | B), P(A₂ | B), and P(A3 | B). (Round your answers to two decimal places.) Events P(A₁) | P(B | A;) P(A¡n B) P(A¡ | B) A₁ 0.20 0.50 A₂ 0.30 0.30 A3 0.50 0.40 1.00 1.00 0.00arrow_forwardA certain system can experience three different types of defects. Let Ai (i = 1,2,3) denote the event that the system has a defect of type i. Suppose that the following probabilities are true. P(A1) = 0.10 P(A2) = 0.07 P(A3) = 0.06P(A1 ∪ A2) = 0.11 P(A1 ∪ A3) = 0.13P(A2 ∪ A3) = 0.11 P(A1 ∩ A2 ∩ A3) = 0.01 (c) Given that the system has at least one type of defect, what is the probability that it has exactly one type of defect? (Round your answer to four decimal places.) (d) Given that the system has both of the first two types of defects, what is the probability that it does not have the third type of defect? (Round your answer to four decimal places.)arrow_forwardS P(A/B): A Let event A = Student belongs to at least one club on campus and event B = Student participates in team sports on campus P(B|A) = ANB At AnyTown Polytechnic College, 42% of students belong to a campus club, 38% of students participate in team sports, and 5% belong to at least one campus club and participate in team sports. Compute the probabilities requested below. Round all probabilities to four decimal places, as needed. P(ANB) = P(AnB): Barrow_forward
- A nuclear reactor becomes unstable if both safety mechanisms A and B fail. The probabilities of failure are P(A) =1/300 and P(B) = 1/200. Also, if A has failed, B is then more likely to fail: P(B/A)=1/100. a) What is the probability of the reactor going unstable? b) If B has failed, what is the probability of the reactor going unstable?arrow_forwardA machine has four components, A, B, C, and D, set up in such a manner that all four parts must work for the machine to work properly. Assume the probability of one part working does not depend on the functionality of any of the other parts. Also assume that the probabilities of the individual parts working are P(A) = P(B) = 0.95, P(C) = 0.91, and P(D) = 0.96. Find the probability that at least one of the four parts will work. Round to six decimal places.arrow_forwardConsider purchasing a system of audio components consisting of a receiver, a pair of speakers, and a CD player. Let A₁ be the event that the receiver functions properly throughout the warranty period, A₂ be the event that the speakers function properly throughout the warranty period, and A3 be the event that the CD player functions properly throughout the warranty period. Suppose that these events are (mutually) independent with P(A₁) = 0.92, P(A₂) = 0.96, and P(A3) = 0.90. (Round your answers to four decimal places.) (a) What is the probability that all three components function properly throughout the warranty period? (b) What is the probability that at least one component needs service during the warranty period? (c) What is the probability that all three components need service during the warranty period? (d) What is the probability that only the receiver needs service during the warranty period? (e) What is the probability that exactly one of the three components needs service…arrow_forward
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