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

Answer: (A)

54

Answer 2

Given data:

| Plant A| Plant B
---|---|---
Manufactured| $60\%$| $40\%$
Standard quality| $80\%$| $90\%$

Define:

• A : Event that the motorcycle is of standard quality.
• B : Event that the motorcycle was manufactured at plant B.
• C : Event that the motorcycle was manufactured at plant A.

The probabilities are:

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

The conditional probabilities are:

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

Using Bayes’ theorem:

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

Substitute the values:

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

Simplify:

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

Now:

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

Final Answer: $126p = 54.$