Question

A company has two plants A and B to manufacture motorcycles. 60% motorcycles are manufactured at plant A and the remaining are manufactured at plant B. 80% of the motorcycles manufactured at plant A are rated of the standard quality, while 90% of the motorcycles manufactured at plant B are rated of the standard quality. A motorcycle picked up randomly from the total production is found to be of the standard quality. If p is the probability that it was manufactured at plant B, then 126p is

• A. 54
• B. 64
• C. 66
• D. 56

Answer 1

The Correct Option is A

Data provided:

	Plant A	Plant B
Manufactured	\(60\%\)	$40\%$
Standard quality	$80\%$	$90\%$

Definitions:

• A: Motorcycle is of standard quality.
• B: Motorcycle manufactured at plant B.
• C: Motorcycle manufactured at plant A.

Probabilities:

\( P(C) = \frac{60}{100}, \quad P(B) = \frac{40}{100}. \)

Conditional probabilities:

\( P(A \mid C) = \frac{80}{100}, \quad P(A \mid B) = \frac{90}{100}. \)

Applying Bayes’ theorem:

\[ P(B \mid A) = \frac{P(A \mid B) P(B)}{P(A \mid B) P(B) + P(A \mid C) P(C)}. \]

Substituting values:

\[ P(B \mid A) = \frac{\frac{90}{100} \times \frac{40}{100}}{\frac{90}{100} \times \frac{40}{100} + \frac{80}{100} \times \frac{60}{100}}. \]

Simplification:

\[ P(B \mid A) = \frac{90 \times 40}{90 \times 40 + 80 \times 60} = \frac{3600}{3600 + 4800} = \frac{3600}{8400} = \frac{3}{7}. \]

Calculation:

\[ 126p = 126 \times \frac{3}{7} = 54. \]

Final Answer: \( 126p = 54. \)