According to Bayes' Theorem, the probability of event A, given that event B has occurred, is as follows.P(A) P(BA)P(A|B)=P(A) P(BA)+P(A').P(B| A')Use Bayes' Theorem to find P(A|B) using the probabilities shown below.1P(A) = P(A) = P(B| A)= ½, and P (B| A') = 2The probability of event A, given that event B has occurred, is P(A| B) = ☐.(Round to the nearest thousandth as needed.)

Bayes theorem of probability can be used to calculate conditional probability. It is the…

Answered: According to Bayes' Theorem, the…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

See similar textbooks

Similar questions

S 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): B
A student goes to the library. Let event A be that the student checks out a book and event B be that the student uses a library computer. Suppose that we happen to know that P (A) = 0.40, that P (B) = 0.30, and that P (B|A) = 0.5. Compute the following (1) P(Ac)(2) P(A AND B) (3) P(A|B)
P(A) + P(B) = 1 for any events A and B that are mutually exclusive. True False
K The probability of event A, given that event B has occurred, can be found using Bayes's Theorem. P(A) P(BA) P(A) P(B) A) + P(A) -P(BA) P(A| B)=- Use Bayes's Theorem to find P(A| B) using the probabilities shown below. P(A) = 0.25, P (A') = 0.75, P(B| A) = 0.1, and P(B| A') = = 0.5 The probability of event A, given that event B has occurred, is (Round to the nearest thousandth as needed.)
A branch of a certain bank has six ATMs. Let X represent the number of machines in use at a particular time of day. The cdf of X is as follows: F(x) = (b) Calculate the following probabilities directly from the cdf: (a) p(2), that is, P(X = 2) (c) 0 0.09 0.21 0.33 0.53 0.90 0.99 1 P(X > 3) x < 0 0 < x < 1 1 ≤ x < 2 2
I know that P(Bc | A) = [P(A | Bc) x P(Bc)] / P(A) but I cannot figure out what P(A) is.
According to Bayes' Theorem, the probability of event A, given that event B has occurred, is as follows. P(A) P(BA) P(A) P(BA) + P(A) P(BA) Use Bayes' Theorem to find P(A| B) using the probabilities shown below. P(A| B)= 1 1 P(A) = 6' ) = -5₁₁ P (A) = ₁ P(B| A) = 77₁ and P (B| A') == 6' 2 The probability of event A, given that event B has occurred, is P(A| B) = (Round to the nearest thousandth as needed.)
Possible values of X, the number of components in a system submitted for repair that must be replaced, are 1, 2, 3, and 4 with corresponding probabilities 0.35, 0.15, 0.35, and 0.15, respectively. (a) Calculate E(X) and then E(5 - X). E(X) E(5-X) = (b) Would the repair facility be better off charging a flat fee of $70 or else the amount $ 150 (5-X) The repair facility ---Select--- be better off charging a flat fee of $70 because E ]= 150 (5-X)] ? Note: It is not generally true that : E ( =) = E(Y)
Let P(E) 0.45, P(F) = 0.6, and P(Fn E) 0.3. Draw a Venn diagram and find the conditional probabilities. P(E | FC (a) P(F | EC) (b)
P(A) = 0.37, P(B) = 0.20, and P(A∩B) = 0.074. Are events A and B independent? Justify the answer mathematically. No, because 0.37 + 0.20 ≠ 0.074 Yes, because (0.37)(0.20) = 0.074 No, because (0.37)(0.20) = 0.074 Yes, because 0.37 + 0.20 ≠ 0.074

Recommended textbooks for you

A First Course in Probability (10th Edition)

Probability

ISBN:

9780134753119

Author:

Sheldon Ross

Publisher:

PEARSON

A First Course in Probability

Probability

ISBN:

9780321794772

Author:

Sheldon Ross

Publisher:

PEARSON

A First Course in Probability (10th Edition)

Probability

ISBN:

9780134753119

Author:

Sheldon Ross

Publisher:

PEARSON

A First Course in Probability

Probability

ISBN:

9780321794772

Author:

Sheldon Ross

Publisher:

PEARSON

SEE MORE TEXTBOOKS