A sample space consists of 9 elementary outcomes E₁, E2, ...,E, whose probabilities are:P(E₁) = P(E₂) = 0.09, P(E3) = P(E4) = P(E5) = 0.1,P(E6)=P(E₂) = 0.2, P(Eg) = P(E₂) = 0.06A = {E₁, E5, Eg}, B = {E2, E5, Eg, E9}, then(a) Calculate P(A), P(B), and P(AB).If(b) Using the addition law of probability, calculate P (AUB).(c) List the composition of the event A U B, and calculateP (AUB) by adding the probabilities of the elementaryoutcomes.(d) Calculate P (B) from P (B), also calculate P (B) directlyfrom the elementary outcomes of B.

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Answered: A sample space consists of 9 elementary…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Assumed a source produces only two digits ("1" and "0") with probabilities of 0.6 and 0.4, respectively. (1) What probability will two and three zeros occur in a five-digit sequence. (2) What probability will at least three ones occur in a five-digit sequence? O (1) 0.230. (2) 0.683. O (1) 0.320. (2) 0.638. (1) 0.23. (2) 0.638. O (1) 0.320. (2) 0.683. O Other:
A gambler simulates 20 plays of a card game, wins 1 time, and estimates P(winning) = 1/20 = 0.05 He then simulates 5,000 plays of the game, wins 450 times, and estimates P(winning) = 450/5,000 = 0.09 Which estimate, 0.05 or 0.09 is probably closer to the true theoretical probability and why? A. 0.05 because it's a smaller number B. 0.09 because the Law of Large Numbers says that the larger the number trials, the better the estimate of the probability C. 0.05 because the Law of Small Numbers says that the smaller the number trials, the better the estimate of the probability D. Both are equally close
6. There are two factories that make electric switches: Morris (M) and Grant (G.) Only 25% of the switches are made in the Morris factory. Of the switches made by Morris, 10% failed inspection. Of the switches made by Grant, 20% failed inspection. Use F for "failed inspection and FC. for "did not fail inspection.." Round your answers to 3 decimal places. Calculate the following probabilities: a. P(M) = P(G)= P(F|M)= P( F°|M)= P(F|G)= P(F°|G) = b. Draw a tree diagram and place the probabilities on the appropriate branch. Insert a screen shot of your tree diagram in your HW PDF. c. Calculate the following probabilities. P(F N G) = P(F' N G) = P(F N M)= P(F' N M) = For the following questions, Do not just give a number as your answer. Use the appropriate probability notation. d. Probability that a switch failed inspection: e. Probability that a switch that failed inspection was made by the Morris factory.
Miami Beach Medical Clinic (MBMC) has about 17.5 patients arriving per hour on average. MBMC serves about 20 patients per hour on average. Also, MBMC does NOT follow the traditional M/M/1 model. A patient arriving at this facility is expected to have one of the following wait times of (0 or 1 or 4 or 6 or 8 minutes) in the queue before being served. The probabilities associated with these wait times in the queue are 0.3, 0.2, 0.3, 0.1, and 0.1 respectively. The patient’s average wait time in queue is expected to be: a. 2.3 minutes b. 7.2 minutes c. 5.6 minutes d. 3.9 minutes e. 2.8 minutes
Q3] on the basis of past experience, the probability that a certain electrical component will be satisfactory is 0.98 the components are sampled item by item from continuous production .in sample of five component, what are the probabilities of finding.0f Two or more defective
The probabilities of events E, F, and EnF are given below. Find (a) P(EIF), (b) P(FIE), (c) P (EIF'), and (d) P (FIE'). 1 P(E) = ₁ P(F) = —, P(ENF) = =1/11 a. P(EIF) = (Type an integer or a simplfied fraction.) b. P(FIE) = (Type an integer or a simplfied fraction.) C. P (EIF') = (Type an integer or a simplified fraction.) d. P (FIE') = (Type an integer or a simplfied fraction.)
Suppose A and B are two events with probabilities: P(A°) = .25, P(B) = .35, P(AN B) = .30. a) What is (A|B) ? b) What is (B|A) ?
An experiment can result in one of five equally likely simple events, E₁, E₂, A: E₂, E4 2' P(A) = = 0.4 B: E₁, E3, E4, E5 P(B) = 0.8 C: E₂, E3 Refer to the following probabilities. P(A n B) = 0.2 P(AIB) = 0.25 P(BIA) P(BU C) = 1 P(BIC) 0.5 P(CIB) 0.25 Use the Addition and Multiplication Rules to find the following probabilities. (a) P(AUB) (b) P(An B) = 0.5 P(A U B) = ---Select--- + ---Select--- ✓ P(A n B) = ---Select--- (c) P(B n C) Yes P(B n C) = ---Select--- No P(C) = 0.4 . P(B) = = P(C) = = ---Select--- V = Do the results agree with those obtained by listing the simple events in each? P(A U B) = P({E₁, E₂, E3, E4, E5}) = 1 P(A n B) = P({E₁}) = 0.2 P(B n C) = P({E3}) = 0.2 .., E5. Events A, B, and C are defined as follows.
56.6
For Part B. Options are Excellent, Good, Fair for each blank.
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?
5. A decision maker subjectively assigned the following probabilities to the all the four possible outcomes of an activity: P(E1) = 0.35, P(E2) = 0.12, P(E3) = 0.44, and P(E4) = 0.20. (a) Are these probability assignments valid? Yes or No. (b) Give reason for your answer in 5(a).

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