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

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: \( 126p = 54. \)

Answer 2

To solve this problem, we'll use Bayes' Theorem, which helps in finding the probability of an event based on prior knowledge of conditions related to the event. Let's define the events as follows:

• \(A_1\): A motorcycle is manufactured at Plant A.
• \(A_2\): A motorcycle is manufactured at Plant B.
• \(B\): A motorcycle is of standard quality.

We need to find the probability \(P(A_2 | B)\), the probability that a motorcycle was manufactured at Plant B given that it is of standard quality.

The given probabilities are:

• \(P(A_1) = 0.6\)
• \(P(A_2) = 0.4\)
• \(P(B | A_1) = 0.8\) (80% of motorcycles from Plant A are of standard quality)
• \(P(B | A_2) = 0.9\) (90% of motorcycles from Plant B are of standard quality)

Using Bayes' theorem:

\(P(A_2 | B) = \frac{P(B | A_2) \cdot P(A_2)}{P(B)}\)

To find \(P(B)\), we use the law of total probability:

\(P(B) = P(B | A_1) \cdot P(A_1) + P(B | A_2) \cdot P(A_2)\)

Plugging in the values:

\(P(B) = 0.8 \cdot 0.6 + 0.9 \cdot 0.4 = 0.48 + 0.36 = 0.84\)

Now we can calculate \(P(A_2 | B)\):

\(P(A_2 | B) = \frac{0.9 \cdot 0.4}{0.84} = \frac{0.36}{0.84}\)

Simplifying the fraction:

\(P(A_2 | B) = \frac{9}{21} = \frac{3}{7}\)

According to the problem, we need to find \(126p\), where \(p = P(A_2 | B)\):

\(126p = 126 \cdot \frac{3}{7} = 54\)

Therefore, the correct answer is 54.