Different illnesses can produce identical symptoms. Suppose a particular set of symptoms, which we will denote as event H,occurs only when any one of three illnesses-A, B, or C-occurs. (For the sake of simplicity, we will assume that illnesses A, B, andC are mutually exclusive.) Studies show the following probabilities of getting the three illnesses.P(A) = 0.09P(B) = 0.005P(C) = 0.07The probabilities of developing the symptoms H, given a specific illness, are as follows.P(HIA) 0.65P(HIB) = 0.95P(HIC) 0.85Assuming that an ill person shows the symptoms H, what is the probability that the person has illness A? (Round your answer tofour decimal places.)

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Answered: Different illnesses can produce…

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...

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If ?(? ∪ ?) = 0.64 and Event A and B are mutually exclusive, which of the following could be possible probabilities for A and B? a. ?(?)=0.64 , ?(?)=0.64 b. ?(?)=0.8 , ?(?)=0.8 c. ?(?)=0.42 , ?(?)=0.22 d. ?(?)=0.8 , ?(?)=0.16
(i) If volume is high this week, then next week it will be high with a probability of 0.9 and low with a probability of 0.1.(ii) If volume is low this week then it will be high next week with a probability of 0.5.Assume that state 1 is high volume and that state 2 is low volume. 1. If the volume this week is high, what is the probability that the volume will be high two weeks from now? 2. What is the probability that volume will be high for three consecutive weeks?
4. A merchant stocks a certain perishable item. He knows that on any given day, he will have a demand for either 2, 3, or 4 of this items with probabilities 0.1, 0.4 and 0.5 respectively. What is the average daily demand?
. Initial diagnosis of a patient’s illness was potentially flu (F), cold (C), or hay fever (H), with probabilities 0.1, 0.2, and 0.7 respectively. A lab test was then taken from the patient. Past tests on similar patients show the following conditional probabilities on the test result being positive (+), neutral (0), or negative (-):•P(+|F) = 0.7P(0|F) = 0.2P(-|F) = 0.1•P(+|C) = 0.1P(0|C) = 0.8P(-|C) = 0.1•P(+|H) = 0.1P(0|H) = 0.2P(-|H) = 0.7•Use the Bayes Theorem to estimate the respective probabilities of F, C, and H given that a +, 0, or – test result has been observed.
Question 2: A. A table of joint probabilities is shown below. A1 A2 Аз B1 0.15 0.25 0.20 B2 0.10 0.15 0.15 Page 1 of 6 a. Calculate the marginal probabilities of event A b. Calculate the marginal probabilities of event B c. Calculate P(A1 | B1) d. Calculate P(B1 | A1) e. Calculate P(B1 | B2) f. Calculate P(A1 | A3) g. Calculate P(A1 U B1)
Suppose we have events A, B, and C with probabilities: P ( A ) = 0.4 , P ( B ) = 0.2 , P ( C ) = 0.3. We also know that: A and B are independent, A and C are mutually exclusive, and P ( B ∩ C ) = 0.15 Compute each of the following. Write your answers as decimals. P ( A ∩ B ) = P ( A ∪ B ) = P ( B ∪ C ′ ) =
Problem 2 [12 points]: Fourth digit of ID a. Genuine motor company manufactures an assembly which consists of three mechanical components. Suppose that the probability of not meeting specification for the first component is equal to 0.09/ 0.11/ 0.13. For the second and third components the probabilities of not meeting specifications are 0.05/ 0.04/ 0.03 and 0.06/ 0.07/ 0.08 respectively. Assuming that the components are independent in terms of not meeting the specification, determine the probability mass function of the number of components in the assembly that does not meet specification. (6.0) Fifth digit of ID Last digit of ID
Question 1 Events A, B and C are disjoint. For the following event probabilities: P(A)=0.16, P(B)=0.69, P(C)=0.15, P(D|A)=0.072, P(D|B)=0.033, P(D|C)=0.061, calculate P(B|D). Your answer: 0.178 0.595 0.860 0.358 0.524 0.341 0.652 0.999 0.161 0.188
An experiment consists of 5 elementary outcomes, with the following probabilities: e | Pr(e) e1 | 0.1 e2 | 0.1e3 | 0.2e4 | 0.3e5 | 0.3 If E = {e1,e2,e3} and F = {e2,e3,e4}, then Pr(E and F) isa. .50 b. .70 c. .67 d. .30
An experiment can result in one of five equally likely simple events, E1, E2, , E5. Events A, B, and C are defined as follows. A: E1, E5 P(A) = 0.4 B: E2, E3, E4, E5 P(B) = 0.8 C: E1, E2 P(C) = 0.4 Refer to the following probabilities. P(A ∩ B) = 0.2 P(A|B) = 0.25 P(B|A) = 0.5 P(B ∪ C) = 1 P(B|C) = 0.5 P(C|B) = 0.25 Use the Addition and Multiplication Rules to find the following probabilities. (a) P(A ∪ B) P(A ∪ B) = + − = (b) P(A ∩ B) P(A ∩ B) = · P(B) = (c) P(B ∩ C) P(B ∩ C) = · P(C) = Do the results agree with those obtained by listing the simple events in each? P(A ∪ B) = P({E1, E2, E3, E4, E5}) = 1 P(A ∩ B) = P({E5}) = 0.2 P(B ∩ C) = P({E2}) = 0.2 YesNo An experiment can result in one or both of the events A and B, with the probabilities shown below. Use the probability table to answer this question. A AC B 0.37…

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