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

P(X=0)=1x10^-5

P(X=1)=1.67x10^-3

P(X=2)=0.0766

P(X=3)=0.9217

Step-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 independents

P(A)=0.95

P(B)=0.98

P(C)=0.99

P(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^-5

P(X=0)=1x10^-5

P(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^-3

P(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)=0.0766

P(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

number one is the correct answer

step-by-step explanation: dogs and cats are mentioned, no other animals

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47