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?

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)?

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

you didn't post any answer choices, so its not 100% clear what your teacher is looking for

however, log(15/7) is equivalent to log(15) - log(7)

i'm using the rule log(x/y) = log(x) - log(y)