a x Mail Davion a er X Homework 9 Mar 1 x \\G chegg Study Guidex 2 Rounding Numb
ID: 3143246 • Letter: A
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a x Mail Davion a er X Homework 9 Mar 1 x G chegg Study Guidex 2 Rounding Number: NC Probability Calculatc x sk wyzant Resou x G Atray of electronic x A: x > C www.webassign net/web Student Assignment Responses/submit?pos:68 dep= 16345902 Homework 9 Math115 Summer 17 (Homework) Instructor Sami Shahin Current Score : 2.48 120 Due : Wednesday, July 26 2017 11:00 PM CDT Print Assignment Question 1 2 34 567 89 10 11 12 13 14 1516 17 1819 20 Total Points Description Assignment Submission For this assignment, you submit answers by questions. Assignment Scoring Over Chapter 7 Your best submission for each entire question is used tor your score 1. 011 points I Previous Answers RolfFM3 7.2.021. 215 Submissions Used My Notes A tray of electronic components contalns 24 components, 5 of which are defective. If 4 components are selected, what Is the possiblity of each of the following? (Round your answers to five decimal places (a) that all 4 are defective (b) that 3 are defective and 1 is good (c) that exactly 2 are defective (d) that none are defective Blurnan 5th Chap doc ShowallXExplanation / Answer
Possibility or probability can be calculated by the ratio of number of ways which satisfies the condition to the number of total possible ways.
a) all four are defective
Number of possible ways = number of ways of selecting 4 components from 5 defective components that is C(5,4) where C represents the combination .
Total number of ways of selecting 4 components from 24 components = C(24,4)
Possibility = C(5,4)/C(24,4)=5*4*3*2*1/(24*23*22*21)
Hence possibility = 0.00047
b)3 are defective and 1 is good
Number of possible ways = 3 components selected from 5 components and 1 component is selected from 19 good components = C(5,3)*C(19,1)
Possibility = C(5,3)*C(19,1)/C(24,4) = 5*4*19*4*3*2*1/(24*23*22*21)
Possibility= 0.03756
C) 2 are defective
Number of possible ways = 2 components from 5 defective components and 2 components from 19 good>
Possibility= C(5,2)*C(19,2)/C(24,4)= 5*4*19*18*4*3*2*1/(24*23*22*21)
Possibilty = 0.64370
D) none of them are defective
Number of possible ways = 4 components should be selected from 19 good ones . no defective component should be selected = C(19,4)
Possibilty = C(19,4)/C(24,4)= 19*18*17*16/(24*23*22*21)
possibility =0.36476.
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