Question

In a bolt factory, machines \(A\), \(B\), and \(C\) manufacture \(25\%\), \(35\%\), and \(40\%\) of the total output respectively. There is a chance of having \(5\%\), \(4\%\), and \(2\%\) defective bolts manufactured by \(A\), \(B\), and \(C\) respectively. If a bolt is drawn at random from the output, then the probability that it is defective is:

• A. \( \dfrac{69}{2000} \)
• B. \( \dfrac{59}{2000} \)
• C. \( \dfrac{79}{2000} \)
• D. \( \dfrac{89}{2000} \)

Hint:

For manufacturing and defective-item problems: \[ P(D)=\sum P(\text{Machine})\times P(\text{Defect from machine}) \] This is a direct application of the theorem of total probability.

Answer 1

The Correct Option is A

Step 1: Set up total probability.
A bolt is defective through one of three machines, so $P(D)=P(A)P(D|A)+P(B)P(D|B)+P(C)P(D|C)$.
Step 2: Convert percentages to decimals.
Shares: $P(A)=0.25$, $P(B)=0.35$, $P(C)=0.40$. Defect rates: $P(D|A)=0.05$, $P(D|B)=0.04$, $P(D|C)=0.02$.
Step 3: Multiply each pair.
$0.25\times0.05=0.0125$, $0.35\times0.04=0.014$, $0.40\times0.02=0.008$.
Step 4: Add them.
$0.0125+0.014+0.008=0.0345$.
Step 5: Turn into a fraction.
$0.0345=\dfrac{345}{10000}$.
Step 6: Reduce.
Divide top and bottom by $5$: $\dfrac{69}{2000}$. \[ \boxed{\frac{69}{2000}} \]