An assembly consists of three mechanical components. Suppose that the probabilities that the first, second, and third components meet specifications are 0.95, 0.98, and 0.99, respectively. Assume that the components are independent. Determine the probability mass function of the number of components in the assembly that meet specifications. 3.1.18 The distribution of the time until a Web

Answer:P(X=0)=1x10^-5P(X=1)=1.67x10^-3P(X=2)=0.0766P(X=3)=0.9217Step-by-step explanation:X="The number of components that meet specifications"A="The 1st component meet the specification"B="The 2nd component meet the specification"C="The 3rd component meet the specification"The events are independentsP(A)=0.95P(B)=0.98P(C)=0.99P(X=0)=P(A'∩B'∩C')=P(A')P(B')P(C')=(1-P(A))(1-P(B))(1-P(C))= (1-0.95)(1-0.98)(1-0.99)=0.05x0.02x0.01=1x10^-5P(X=0)=1x10^-5P(X=1)=P(A∩B'∩C')+P(A'∩B∩C')+P(A'∩B'∩C)P(X=1)=P(A)P(B')P(C')+P(A')P(B)P(C')+P(A')P(B')P(C) P(X=1)=P(A)(1-P(B))(1-P(C))+(1-P(A))P(B)(1-P(C))+(1-P(A))(1-P(B))P(C)P(X=1)=(0.95)(1-0.98)(1-0.99)+(1-0.95)(0.98)(1-0.99)+(1-0.95)(1-0.98)(0.99) P(X=1)=1.67x10^-3P(X=2)=P(A'∩B∩C)+P(A∩B'∩C)+P(A∩B∩C')P(X=2)=P(A')P(B)P(C)+P(A)P(B')P(C)+P(A)P(B)P(C') P(X=2)=(1-P(A))P(B)P(C))+P(A)(1-P(B))P(C)+P(A)P(B)(1-P(C))P(X=2)=(0.05)(0.98)(0.99)+(0.95)(0.02)(0.99)+(0.95)(0.98)(0.01) P(X=2)=0.0766P(X=3)=P(A∩B∩C)=P(A)P(B)P(C)=P(A)P(B)P(C)= (0.95)(0.98)(0.99)=P(X=3)=0.9217