Question

Components arriving at a distributor are checked for defects by two different inspectors (each component is checked by both inspectors). The first inspector detects 97% of all defectives that are present, and the second inspector does likewise. At least one inspector does not detect a defect on 6% of all defective components. What is the probability that the following occur?

Answer 1

Explanation:

The probability for first inspector to detect defective, P(A) = 0.97

The probability of second inspector to detect defective, P(B) = 0.97

Then, the probability that at least one inspector does not detect all defective components, P(C) = 0.06

The probability that the first inspector detected a defect and could not detect all defective components = P(A)*P(C) = 0.97 X 0.06 = 0.0582

The probability that the second inspector detected a defect and could not detect all defective components = P(B)*P(C) = 0.97 X 0.06 = 0.0582

The probability that at least one inspector detected all defective components, P(D) = 1 - 0.06 = 0.9994

The probability that the first inspector detected a defect and also detected all defective components = P(A)*P(D) = 0.97 X 0.9994 = 0.969

The probability that the second inspector detected a defect and also detected all defective components = P(B)*P(D) = 0.97 X 0.9994 = 0.969

Answer 2

What is the probability that the following occur?

a. A defective component will be detected only by the first inspector? By exactly one of the two inspectors?

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

Answer:

a

I. 0.291

Ii. 0.582

b.

0.0027

Step-by-step explanation:

Given

Probability that an inspection is right = P(R) = 0.97

Probability that an inspection is wrong = P(W) = 1 - P(R) = 1 - 0.97 = 0.03

a.

I. A defective component will be detected only by the first inspector?

This means that the first inspection is right and the second is wrong

i.e. P(R) * P(W)

Since inspectors are independent of each other, solution is:

= 0.97 *0.03

= 0.291

Ii. By exactly one of the two inspectors?

There are two ways of achieving this

This means that either first inspection is right and second is wrong or first inspection is wrong and second inspection is right.

Since inspectors are independent of each other, solution is:

P(R) * P(W) + P(W) *P(R)

= 0.97 * 0.03 + 0.03 *0.97

= 2 * 0.97 * 0.04

= 0.582

b. All three defective components in a batch escape detection by both inspectors..

This means that both inspection are wrong three times

Since the inspection are independent, the solution is

P(W) * P(W) or P(W) * P(W) or P(W) * P(W)

= P(W) * P(W) * 3

= 0.03 * 0.03 * 3

= 0.0027