Events A₁ and A₂ are mutually exclusive and form a completepartition of a sample space S with P (A₁) = 0.41. If E is anevent in S with P (E | A₁) =0.13 and P(E|A₂) = 0.12,compute P (E) =? (Hint: Because A₁ and A2 are mutuallyexclusive and form a complete partition of the sample space,P(A₂) = 1- P(A₁)).Note: If your final answer has up to four decimal places, please enter yourfull answer without rounding it. If your answer contains more than fourdecimal places, please round it to four decimal places before entering it inthe box below.

It is given that, A1 and A2 are mutually exclusive events and form a complete partition of the…

Answered: Events A₁ and A2 are mutually exclusive…

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

7. Let A and B be two events from the same sample space with P(A) = 0.35 and P(B) = 0.70. (a) Is it possible that the events A and B are mutually exclusive? Fully justify your answer. (b) Suppose that P(A|B) = 0.2, find P(An Bc) (fully justify your answer analytically). (c) Is it possible that P(A|B) = 0.60? Justify your answer analytically.
An analysis of personal loans at a local bank revealed the following facts: 10% of all personal loans are in default (D), 90% of all personal loans are not in default (D'), 20% of those in default are homeowners (H | D), and 70% of those not in default are homeowners (H | D'). If a personal loan is selected at random, P (D | H) = ___. O 0.78 0.02 0.03 0.18 O 0.63
No.1
Let A, B and C be events from a common sample space such that: P(A) = 0.72, P(B) = 0.56, P(C) =0.25, P(ANB) = 0.42. P(ANC) = 0.18, P(BOC) =0 %3D Compute P(AUBUC).
Let S be a sample space and E and F be events associated with S. Suppose that Pr(E) = 0.5, Pr(F) = 0.3 and Pr(ENF) = 0.2. Calculate the following probabilities. (a) Pr(E|F) (b) Pr(F|E) (c) Pr(티F') (d) Pr (E'JF') ... (a) To find Pr(E|F), simplify the fraction with Pr(EnF) in the numerator and Pr(F) in the denominator. Pr(E|F) = || |(Type an integer or a simplified fraction.)
Look at the 4 probability rules beow , an provide a REAL LIFE example for each. Also, with a Real LIfe EXAMPLE, explain how Disjoint events CAN be dependent. Rule 1. The probability P(A) of any event A satisfies 0 ≤ P(A) ≤ 1. Rule 2. If S is the sample space in a probability model, then P(S) = 1. Rule 3. Two events A and B are disjoint if they have no outcomes in common and so can never occur together. If A and B are disjoint, P(A or B) = P(A) + P(B) This is the addition rule for disjoint events. Rule 4. For any event A, P(A does not occur) = 1 − P(A)
Suppose we have a sample with five equally likely experimental outcomes: E1, E2, E3, E4, E5 Let: A = (E1, E3, E4, E5) B= (E1, E3} C = (E2, E3, E4,) a) There are 3 methods of assigning probabilities - classical, subjective and relative frequency. Which method is used above? b) Find P(A), P(B) and P(C) c) Find A and P(AG) d) Find P(B U C^C) e) Are A and C mutually exclusive? Explain.
The finish of a hole depends on the edge of the bit used, only 1% of the holes with a new drill have an irregular cut, 3% have an irregular cut with average bits and 5% present irregular cut with worn bits. If 25% of the holes are made with new bits, 60% have average edges and 15% are worn out. [use 3 decimal places] a) If a piece is taken at random, what is the probability of having holes irregular? b) If a piece with irregular holes is taken, what is the probability that would have been made with an average drill bit?
A sample space consists of just three outcomes, denoted O1, O2, and O3. Events E, F, and G are defined as E = {O1,O2 } F = {O2,O3 } G = {O1,03 } If Pr[E] = 0.77 and Pr[F] = 0.52, what is Pr[G]? %3| %3D (A) 0.645 (B) 0.77 (C) 0.355 (D) 0.29 (E) 0.23 (F) 0.48 (G) 0.52 (H) 0.71
F, G, H, and Y are events in the same sample space with P(A) = 0.26, P(B) = 0.39, and P(C) = 0.18, . F G and H are mutually exclusive. In addition, suppose that P(Y|F) = 0.21, P(Y|G) = 0.43, and P(Y|H) = 0.31. What is the value of P(Y)?

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