Respuesta :
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)?
Answer:
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
