The prior probabilities for events A1, A2, and A3 are P(A1) = 0.10, P(A2) = 0.60, and P(A3) = 0.30.The conditional probabilities of event B given A1, A2, and A3 are P(B|A1) = 0.50, P(B|A2) = 0.40, andР(B Аз) 3D 0.20.Round your answers to two decimal places.a. Compute P(B n A1) , P(BN A2), and P(BN A3).P(BN A1) =0.05P(BN A2) =0.24P(BN A3) =0.06b. Apply Bayes' theorem, to compute the posterior probability P(A2|B).P(A; N B ) =P(A;)P(B|A;)P(A1)P(B|A1 )+ P(A2)P(B|A2 )+...+ P(A,)P(B|A,)c. Use the tabular approach to applying Bayes' theorem to compute P(A1|B), P(A2|B), and P(A3|B).P(A; N B)EventsP(A;)P(B|A;)P(A;|B)A1A2A3Total:

Answered: The prior probabilities for events A1,…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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You have three red cards numbered 1,2,3 and two orange cards numbered 1,2. Finding the following probabilities
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 0.00
The probability of an integrated circuit passing a quality test is F. In a batch of 10 integrated circuits find the probabilities thati. 2 will failii. 8 will passiii. at least 2 will failiv. fewer than 2 will fail.
A 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.
From a large survey of customers using a chain of coffee shops, 60% of the customers are male, 50% purchase food, 15% are both Male and purchase food. What are the probabilities that: A customer is both female and purchases food? A female customer purchases food? A customer purchasing food is female?
3. For two independent events, A and B, P(A)=D0.1 and P(B)-D0.5. Enter your probabilities as decimals. 3(a) P(A and B) = 3(b) P(A| B) = %3! 3(c) P(A or B) =
2) Events A and B have respective probabilities P(A) = 0.4 and P(B) = 0.5.You also know that P(A ∪ B) = 0.7.(a) What is P(A|B)?(b) What is P(B|A)?
1. A and B are independent, with P (A) = 0.6, and P (B) = 0.4, calculate the following probabilities: P (A and B1) b. If A and B are independent, with P (A) = 0.6, and P (B) = 0.4, calculate the following probabilities: P (A/B) c.If A and B are independent, with P (A) = 0.6, and P (B) = 0.4, calculate the following probabilities: P (A1 and B) d.If A and B are independent, with P (A) = 0.6, and P (B) = 0.4, calculate the following probabilities: P (A or B)
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%…
Write the m.g.f. of a r.v. X taking on values -4, 0 and 4 with respective probabilities 0.3, 0.4, and 0.3. 0.3 e4 +0.4 • e² + 0.3 -4z 0.3 e +0.4+ 0.3. e4z 1+0.3. e4z +0.3 · e-4z e4 0.3 ez +0.4 e² + 0.3. e4z
Suppose you have 3 events A,B,C and exactly one happens and are equally likely. There is another event E and we know P(E|A)=.4, P(E|B)=.7, and P(E|C)=.8. Suppose we don't know if A,B, or C was the event that happened, but we do know E happened. Find the three probabilities that A happened, B happened, and C happened.

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