What is the steady-state probability vector?
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Suppose the transition matrix for a Markov process is
State A
State B
State A State B
1
1]
0
1-P р
р
9
where 0 < p < 1. So, for example, if the system is in state A at time 0 then the probability
of being in state B at time 1 is p.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1aa15229-e131-473b-bc97-7b983c6676cc%2F8b4728b3-afdd-4d82-8cdf-276cd7222bf8%2Fxix3oan_processed.png&w=3840&q=75)
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- Don't hand writingA 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?For 10 percent of the products in a category, a firm fails to satisfy all demand during themonth. What is its in-stock probability?
- You are enrolled in MBA 440. At the beginning of the semester, you used your hunch to estimate the likelihood of passing this class. Let P be the event of passing the class; F is the event of failing the class. Your prior probabilities are as follows: P(P)=0.7 P(F)=0.3 After the first week of class, you find out that you have failed the first assignment. Let A1 be the event of passing the first assignment; let A2 be the event of failing the first assignment. Now you feel you need to update your belief about the likelihood of passing this class. You talk to your Professor and she tells you that given someone has passed the class, the probability of passing the first assignment is 0.8, i.e. P(A1|P)=0.8. She also gives you the following conditional probabilities: P(A2|P)=0.2; P(A1|F)=0.3 and P(A2|F)=0.7. What is the posterior probability that you will pass this class given that you have failed the first assignment?IToday's practice The following circuit operates if and only if there is a path of functional devices from the left to the right. Assume the devices fail independently and that the probability of failure for each device is 0.9. What is the probability that the circuit operates? TI 3 T2 4. 5.3
- At the beginning of each day, a patient in the hospital is classifed into one of the three conditions: good, fair or critical. At the beginning of the next day, the patient will either still be in the hospital and be in good, fair or critical condition or will be discharged in one of three conditions: improved, unimproved, or dead. The transistion probabilities for this situation are Good Fair Critical Good 0.65 0.20 0.05 Fair 0.50 0.30 0.12 Critical 0.51 0.25 0.20 Improved Unimproved Dead Good 0.06 0.03 0.01 Fair 0.03 0.02 0.03 Critical 0.01 0.01 0.02 For example a patient who begins the day in fair condition has a 12% chance of being in critical condition the next day and a 3%…An insurance policy on an electrical device pays a benefit of $2400 if the device fails during the first year. The amount of the benefit decreases by $800 each successive year until it reaches 0 . If the device has not failed by the beginning of any given year, the probability of failure during that year is 0.29.Find the expected benefit under this policy.Suppose a life insurance company sells a $280,000 one-year term life insurance policy to a 24-year-old female for $180. The probability that the female survives the year is 0.999522. Compute and interpret the expected value of this policy to the insurance company.
- Willow Brook National Bank operates a drive-up teller window that allows customers to complete bank transactions without getting out of their cars. On weekday mornings, arrivals to the drive-up teller window occur at random, with an arrival rate of 30 customers per hour or 0.5 customers per minute. Let's assume that the service times for the drive-up teller follow an exponential probability distribution with a service rate of 45 customers per hour, or 0.75 customers per minute. Determine the following operating characteristics for the system. (Round your answers to four decimal places.) (a) The probability that no customers are in the system 1353 (b) The average number of customers waiting 1786 (c) The average number of customers in the system 1.25 (d) The average time (in min) a customer spends waiting 4.2 X min (e) The average time (in min) a customer spends in the system x min 1 (f) The probability that arriving customers will have to wait for service 73 XDetermine whether the following are valid probability models or not. Type "VALID" if it is valid, or type "INVALID" if it is not.Eee303 5c
