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:
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:
Show 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.
Updated On: Jun 17, 2026
- \( \dfrac{69}{2000} \)
- \( \dfrac{59}{2000} \)
- \( \dfrac{79}{2000} \)
- \( \dfrac{89}{2000} \)
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The Correct Option is A
Solution and Explanation
Concept:
This problem is based on the theorem of total probability.
If an event can occur through several mutually exclusive cases, then:
\[
P(D)
=
P(A)P(D|A)
+
P(B)P(D|B)
+
P(C)P(D|C)
\]
where:
• \(D\) denotes the event of selecting a defective bolt.
• \(P(D|A)\) means probability of defect given bolt came from machine \(A\).
Step 1: Write all given probabilities in decimal form. Machine contributions: \[ P(A)=25\%=0.25 \] \[ P(B)=35\%=0.35 \] \[ P(C)=40\%=0.40 \] Defective probabilities: \[ P(D|A)=5\%=0.05 \] \[ P(D|B)=4\%=0.04 \] \[ P(D|C)=2\%=0.02 \]
Step 2: Apply total probability theorem. \[ P(D) = P(A)P(D|A) + P(B)P(D|B) + P(C)P(D|C) \] Substituting values: \[ = (0.25)(0.05) + (0.35)(0.04) + (0.40)(0.02) \] \[ = 0.0125+0.014+0.008 \] \[ = 0.0345 \]
Step 3: Convert into fraction form. \[ 0.0345=\frac{345}{10000} \] Dividing numerator and denominator by \(5\): \[ = \frac{69}{2000} \] Therefore, the required probability is: \[ \boxed{\frac{69}{2000}} \]
• \(D\) denotes the event of selecting a defective bolt.
• \(P(D|A)\) means probability of defect given bolt came from machine \(A\).
Step 1: Write all given probabilities in decimal form. Machine contributions: \[ P(A)=25\%=0.25 \] \[ P(B)=35\%=0.35 \] \[ P(C)=40\%=0.40 \] Defective probabilities: \[ P(D|A)=5\%=0.05 \] \[ P(D|B)=4\%=0.04 \] \[ P(D|C)=2\%=0.02 \]
Step 2: Apply total probability theorem. \[ P(D) = P(A)P(D|A) + P(B)P(D|B) + P(C)P(D|C) \] Substituting values: \[ = (0.25)(0.05) + (0.35)(0.04) + (0.40)(0.02) \] \[ = 0.0125+0.014+0.008 \] \[ = 0.0345 \]
Step 3: Convert into fraction form. \[ 0.0345=\frac{345}{10000} \] Dividing numerator and denominator by \(5\): \[ = \frac{69}{2000} \] Therefore, the required probability is: \[ \boxed{\frac{69}{2000}} \]
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