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

the answer is mp

step-by-step explanation:

$2416.6

step-by-step explanation:

per hour = $ 25

total work time = 35

commission = 15%

total she has to make = $ 4,500

let she sold x dollar products

so her commission will be 15 % of x

now

total = weekly basic pay + commission

4500= (25*35) + 15 % of x

4500=875 + * x

4500-875 =   * x

3625= * x

multiplying wwith 100 on both sides gives

36250 = 15 x

now

x =  

x= $2416.6