Components arriving at a distributor are checked for defects by two different inspectors (each component is checked by both inspectors). The first inspector detects 83% of all defectives that are present, and the second inspector does likewise. At least one inspector does not detect a defect on 34% of all defective components. What is the probability that the following occur

Complete question is;

Components arriving at a distributor are checked for defects by two different inspectors (each component is checked by both inspectors). The first inspector detects 83% of all defectives that are present, and the second inspector does likewise. At least one inspector does not detect a defect on 34% of all defective components. What is the probability that the following occurs?

(a) A defective component will be detected only by the first inspector?

b) A defective component will be detected by exactly one of the two inspectors?

(c) All three defective components in a batch escape detection by both inspectors (assuming inspections of different components are independent of one another)?

A) 0.17

B) 0.34

C) 0

Step-by-step explanation:

a) We are told that the first inspector(A) detects 83% of all defectives that are present, and the second inspector(B) also does the same.

This means that;

P(A) = P(B) = 83% = 0.83

We are also told that at least one inspector does not detect a defect on 34% of all defective components.

Thus;

P(A' ⋃ B') = 0.34

Also, we now that;

P(A ⋂ B) = 1 - P(A' ⋃ B')

P(A ⋂ B) = 1 - 0.34

P(A ⋂ B) = 0.66

Probability that A defective component will be detected only by the first inspector is;

P(A ⋂ B') = P(A) - P(A ⋂ B)

P(A ⋂ B') = 0.83 - 0.66

P(A ⋂ B') = 0.17

B) probability that a defective component will be detected by exactly one of the two inspectors is given as;

P(A ⋂ B') + P(A' ⋂ B) = P(A) + P(B) - 2P(A ⋂ B)

P(A) + P(B) - 2P(A ⋂ B) ; 0.83 + 0.83 - 2(0.66) = 0.34

C) Probability that All three defective components in a batch escape detection by both inspectors is written as;

P(A' ⋃ B') - (P(A ⋂ B') + P(A' ⋂ B))

Plugging in the relevant values, we have;

0.34 - 0.34 = 0