Reconsider the system defect situation described in Exercise 26 (Section 2.2). a. Given that the system has a type 1 defect, what is the probability that it has a type 2 defect? b. Given that the system has a type 1 defect, what is the probability that it has all three types of defects? c. Given that the system has at least one type of defect, what is the probability that it has exactly one type of defect? d. Given that the system has both of the first two types of defects, what is the probability that it does not have the third type of defect?

hello exercise 26 is missing attached below is exercise 26

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

the discriminant is 64 + 4*1*12 = 112

this is prime.

step-by-step explanation:

30 feet

step-by-step explanation:

that would be 2 1/2 * 12

= 24 + 6

= 30 feet   (answer)