Chapter 5, Problem T5.12SPT

To determine

(a)

To calculate: The probability that the part is defective.

Answer to Problem T5.12SPT

Probability that a part is defective if a part is selected randomly is 0.19.

Explanation of Solution

Given information:

Percentage of parts produced by Machine A $=60\%$

Percentage of parts produced by Machine B $=\text{ 3}0\%$

Percentage of parts produced by Machine C $=\text{ 1}0\%$

Percentage of parts produced by Machine A are defective $=\text{ 1}0\%$

Percentage of parts produced by Machine B are defective $=\text{ 3}0\%$

Percentage of parts produced by Machine C are defective $=\text{ 4}0\%$

Formula used: As per the law of total probability

  $\text{P (D) = P (D|A) }\times \text{ P(A) + P (D|B) }\times \text{ P (B) + P (D|C) }\times \text{ P (C)}$

Calculation:

Let the probability of parts produced by Machine A be P (A),

The probability of parts produced by Machine B be P (B),

The probability of parts produced by Machine C be P (C),

The probability of defective parts be P (D)

Using the given information,

  $\begin{array}{l}\text{P }\left(\text{A}\right)=0.\text{6}0\\ \text{P }\left(\text{B}\right)=0.\text{3}0\\ \text{P }\left(\text{C}\right)=0.\text{1}0\\ \text{P }\left(\text{D}|\text{A}\right)=0.\text{1}0\\ \text{P }\left(\text{D}|\text{B}\right)=0.\text{3}0\\ \text{P }\left(\text{D}|\text{C}\right)=0.\text{4}0\end{array}$

Applying the above stated theorem,

P (D) $=0.10\times 0.60+0.30\times 0.30+0.40\times 0.10$

P (D) $=0.19$

(b)

To determine

To calculate: The probability that a part is produced by Machine B if the part is found to be defective.

Answer to Problem T5.12SPT

The probability of the part produced by Machine B if found to be defective is 0.47368.

Explanation of Solution

Given information: As per the information in the sub part a.

Formula used: According to Bayes theorem,

  $\begin{array}{l}\text{P (B|D) = }\frac{\text{P (B }\cap \text{ D)}}{\text{P (D)}}\\ \text{P (B }\cap \text{ D) = P (D|B) }\times \text{ P(B)}\end{array}$

Calculation: From the above sub part, P (D) $=0.19$

  $\begin{array}{l}\text{P (B }\cap \text{ D) = 0}\text{.30 }\times \text{ 0}\text{.30 = 0}\text{.09}\\ \text{So, P (B|D) = }\frac{0.09}{0.19}=0.47368\end{array}$