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# HW3sol - IE 300 HW3 SOLUTIONS ASSIGNMENT 2-187 3-27 33 49...

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IE 300 HW3 SOLUTIONS ASSIGNMENT: 2-187, 3-27, 33, 49, 66, 98, 114, 120, 136, 176 from the textbook 5 th edition. 2-187 (a) (30/50)*(29/49)*(28/48)=0.2071 (b) 30/48=0.625 (c) The probability can be written in terms of whether or not the first and second washer selected are thicker than the target. Let A1, A2 and A3 denote the events that the first, second and third washer selected are thicker than the target respectively. Thus P(A3)=P(A1A2A3 or A1A2’A3 or A1’A2A3 or A1’A2’A3) =P(A3|A1A2)P(A1A2)+ P(A3|A1A2’)P(A1A2’) + P(A3|A1’A2)P(A1’A2)+ P(A3|A1’A2’)P(A1’A2’) =P(A3|A1A2)P(A2|A1)P(A1)+ P(A3|A1A2’)P(A2’|A1)P(A1) + P(A3|A1’A2)P(A2|A1’)P(A1’)+ P(A3|A1’A2’)P(A2’|A1’)P(A1’) = g2871g2868 g2873g2868 g2870g2877 g2872g2877 g2870g2876 g2872g2876 + g2871g2868 g2873g2868 g2870g2868 g2872g2877 g2870g2877 g2872g2876 + g2870g2868 g2873g2868 g2871g2868 g2872g2877 g2870g2877 g2872g2876 + g2870g2868 g2873g2868 g2869g2877 g2872g2877 g2871g2868 g2872g2876 =0.6 3-27 Let X denote the number of components that meet specifications; X =0,1,2,3 P(X=0)=P(none of them meet specifications)=0.05*0.02*0.01=0.00001 P(X=1)=P(one of them meets specifications) =0.95*0.02*0.01+0.05*0.98*0.01+0.05*0.02*0.99=0.00167 P(X=2)=P(two of them meet specifications) =0.95*0.98*0.01+0.95*0.02*0.99+0.05*0.98*0.99=0.07663 P(X=3)=P(all of them meet specifications)=0.95*0.98*0.99=0.92169 3-33 From 3-15, we have P(X=-2)=1/8; P(X=-1)=2/8;

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• Spring '09
• Zafarani
• Probability theory, Binomial distribution, Cumulative distribution function, Discrete probability distribution, Negative binomial distribution

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HW3sol - IE 300 HW3 SOLUTIONS ASSIGNMENT 2-187 3-27 33 49...

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