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 ) =
= = 0.5
B) Probability of having all three defects given that it has type 1 defect
= P ( A1 ∩ A2 ∩ A3 | A1 ) =
= = 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