A sample space contains 7 sample points or outcomes, Let events A and B bedefined according to the Venn diagram below.Let P(1) = P(2) = P(3) = P(7) = 0.05,P(4) = P(5) = 0.1,and P(6) = 0.6.Use the Venn diagram and the given probabilities to find:(a) P(An B) =(b) P(AUĀ) =(c) P(B) =(d) P(A) =

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Answered: A sample space contains 7 sample points…

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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Table 3.1 describes the distribution of a random sample S of 100 individuals, organized by gender and whether they are right- or left-handed. Right-handed Left-handed Table 3.1 Left-hand Right- Hand MALE 43 9 FEMALE 44 4 Let's denote the events M = the subject is male, F = the subject is female, R = the subject is right-handed, L= the subject is left-handed. Compute the following probabilities: a. P(M) b. P(F) c. P(R) d. P(L) e. P(M AND R) f. P(F AND L) g. P(M OR F) h. P(M OR R) 1. P(F OR L) J. P(M) k. P(R|M) I. P(F|L) m. P(L|F)
A researcher collected data on adult alcohol consumption and marital status. She reported the number of adults in each category. 1- 60 drinks More than 60 drinks O drinks per month Total per month per month Single 20 100 30 150 Married 100 400 150 650 Widowed 10 25 15 50 Divorced 20 80 50 150 Total 150 605 245 1000 An adult is randomly selected from the study. Use this information to find the probabilities requested. Find the following probabilities as fractions. Don't simplify your results. a) What is the probability that a person drinks 1 - 60 drinks per month given that they are divorced? b) What is the probability that a person is divorced given that they drink 1 - 60 drinks per month? c) What is the probability that a person is single given that they have more than 60 drinks per month? d) What is the probability that a person has more than 60 drinks per month given that they are a widow?
15%0f cellphone users use their cell phones to access internet. In arandom sample of 10 cellphone users, what is probability that exactly 2 have used their phones to access internet?
An experiment can result in one of five equally likely simple events, E,, E, E. Events A, B, and C are defined as follows. A: E,, E, P(A) = 0.4 B: E,, E E E, P(B) = 0.8 C: E,. E, P(C) = 0.4 Refer to the following probabilities. P(AN B) = 0.2 P(A|B) = 0.25 P(B|A) = 0.5 P(B U 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 U B) P(A U B) = --Select-- v+ --Select-- ---Select-- v %3D (b) P(ANB) P(ANB) =-Select-- v P(B) = (c) P(B nC) P(BnC) = -Select-- V- P(C) = Do the results agree with those obtained by listing the simple events in each? P(A U B) = P({E, E, E,, E, E;}) = 1 P(A N B) = P({E,}) = 0.2 P(Bn C) = P({E,}) = 0.2 O Yes O No
In a table of two-digit random numbers, the probabilities of 0.16 to 0.20 can be represented by the numbers: (a) 16 to 20 (b) 0 to 4 (c) 1 to 5 (d) 15 to 19
In a certain school, the probabilities of the number of students who are reported tardy are shown in the following table: Number tardy: 2 3 4 Probability: 0.17 0.25 0.31 0.21 0.06 What is the expected number of tardies (rounded to two decimal places)?
In a sales effectiveness seminar, a group of sales representatives tried two approaches to selling a customer a new automobile: the aggressive approach and the passive approach. For 1160 customers, the following record was kept: Suppose a customer is selected at random from the 1160 participating customers. Let us use the following notation for events: A = aggressive approach, Pa = passive approach, S = sale, N = no sale. So, P(A) is the probability that an aggressive approach was used, and so on. (a) Compute P(S), P(S | A), and P(S | Pa). (Enter your answers as fractions.) Simplify the fraction if needed P(S) = ? P(S | A) = ? P(S | Pa) = 95/116 (b) Compute P(A and S) and P(Pa and S). (Enter your answers as fractions.) Simplify the fraction if needed P(A and S) = ? P(Pa and S) = 95/232 (c) Compute P(N) and P(N | A). (Enter your answers as fractions.) Simplify the fraction if needed
The accompanying tree diagram represents an experiment consisting of two trials. Use the diagram to find the probabilities below. (a) P(A)(b) P(E | A)(c) P(A E)(d) P(E)
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The natural remedy echinacea is reputed to boost the immune system, which will reduce flu and colds. A 6-month study was undertaken to determine whether the remedy works. From this study, the following probability distribution of the number of respiratory infections per year (X) for echinacea users was produced: 3 4 1 2 P(X) 0.342 0.323 0.203 0.074 0.058 Find the following probabilities: A. An echinacea user has more than one infection per year B. An echinacea user has no infection per year C. An echinacea user has between one and three (inclusive) infections per year
An experiment consists of tossing a single die and observing the number of dots that show on the upper face. Use events A, B, and C, defined as follows, to answer the question. A: observe a number less than 5 B: observe a number less than or equal to 3 C: observe a number greater than 4 Refer to the following probabilities. P(S) = 1 P(AIB) P(B) P(An B) = 1 P(BU C) Since ---Select--- P(A n B n C) = 0 1 = = HNO = 1 P(An C) P(B n C) 0 P(AUC) = 1 5 6 = 2 NO 2 0 Are events A and C independent? A and C ---Select--- independent. Are events A and C mutually exclusive? Since ---Select--- A and C ---Select--- mutually exclusive.
If ? and ? are two mutually exclusive events with ?(?)=0.2 and ?(?)=0.7, find the following probabilities: by the way, what's the c and d mean like Both A and B have A horizontal line over their heads

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