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 81% of all defectives that are present, and the second inspector does likewise. At least one inspector does not detect a defect on 38% of all defective components. What is the probability that the following occur? (a) A defective component will be detected only by the first inspector? A defective component will be detected 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 1

Answer:

The probability that defective component will be detected only by the first inspector is 0.19

The probability that defective component will be detected by exactly one of the two inspectors is 0.38

The probability that all three defective components in a batch escape detection by both inspectors is 0.

Step-by-step explanation:

Part (A)

It is given that The first inspector detects 81% of all defectives that are present, and the second inspector does likewise.

Therefore P(A)=P(B)=81%=0.81

At least one inspector does not detect a defect on 38% of all defective components.

Therefore, $P(\overline{A\cap B})=0.38$

As we know:

$P(\overline{A\cap B})=1-P(A\cap B)=0.38\\P(A\cap B)=1-0.38=0.62$

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

$P(A \cap \bar B)=P(A)-P(A\cap B)\\P(A \cap \bar B)=0.81-0.62\\P(A \cap \bar B)=0.19$

The probability that defective component will be detected only by the first inspector is 0.19

Part (B) A defective component will be detected by exactly one of the two inspectors.

This can be written as: $P(A \cap \bar B)+P(\bar A\cap B)$

As we know:

$P(\bar A \cap B)=P(B)-P(A\cap B)\ and\ P(A \cap \bar B)=P(A)-P(A\cap B)$

Substitute the respective values we get.

$P(A \cap \bar B)+P(\bar A\cap B)=P(A)+P(B)-2P(A\cap B)\\=0.81+0.81-2(0.62)\\=1.62-1.24\\=0.38$

The probability that defective component will be detected by exactly one of the two inspectors is 0.38

Part (C) All three defective components in a batch escape detection by both inspectors

This can be written as:$P(\bar A \cup \bar B)-P(\bar A\cap B)-P(A\cap \bar B)$

As we know $P(\overline{A\cap B})=P(\bar A\cup \bar B)=0.38$

From part (B) $P(\bar A\cap B)+P(A\cap \bar B)=0.38$

This can be written as:

$P(\bar A \cup \bar B)-P(\bar A\cap B)-P(A\cap \bar B)=0.38-0.38=0$

The probability that all three defective components in a batch escape detection by both inspectors is 0