Respuesta :
Answer:
hello exercise 26 is missing attached below is exercise 26
answer: a) 0.5 (b) 0.833 (c) 0.357 (d) 0.833
Step-by-step explanation:
A1... A3 = represents defects in system 1 to 3
A' 1 ---- A'3 = represents no defects in systems 1 to 3
P( A1 ∩ A2 ) = 0.06 ( as calculated )
P ( A1 ∩ A3 ) = 0.03 ( as calculated )
P ( A2 ∩ A3 ) = 0.02 ( as calculated )
a) Probability of having a type 2 defect
The probability of A given system having both Type 2 defect given that it has a type 1 defect is considered conditional probability
= p ( A2 | A 1 ) = [tex]\frac{P(A1 n A2)}{P(A1)}[/tex]
= [tex]\frac{0.06}{0.12}[/tex] = 0.5
B) Probability of having all three defects given that it has type 1 defect
= P ( A1 ∩ A2 ∩ A3 | A1 ) = [tex]\frac{P( A1 n A2 n A3 )}{P(A1)}[/tex]
= [tex]\frac{0.01}{0.12}[/tex] = 0.0833 ≈ 0.8
C) probability of having exactly one type of defect given that it has atleast one type of defect
= P ( exactly one Ai | at least one Ai ) = 0.357
attached below is the detailed solution
D) probability of not having the third defect given that it has the first two types of defects
= P ( A'3 | A1 ∩ A2 ) = 0.833
attached below is the detailed solution



The system situation
As per the question, the system defect situation describes the exercise given in section 2.2 which shows the system has a probability of type 1 defect. The type 2 of defect is given in the system of type 1 and the chances of probability that it all has three types of defects.
The answer is 0.5, 0.83 and 035, and 0.83.
- As per the exercise, the defects that can be measured froth type of system that is five on the defects and from the system of type 1 defects.
- The chances of the chances is 0.5 and the given system of the type defects has the probability of 0.5 with
- The third type of defect or the least one type is 0.357. Both of the first types of defects has a probability of 0.833.
Learn more about the system defect situation.
https://brainly.com/question/23692194