A company has two plants A and B to manufacture motorcycles. 60% motorcycles are manufactured at pla

Q: Why can't I just take the ratio of the percentages directly, like 90/(80+90)?

That is a classic trap! You are ignoring the 'weight' of each plant. Plant A produces 60% of the total, while Plant B produces only 40%. You must account for the volume of production from each source before comparing their quality rates. Bayes' Theorem naturally handles these weights through the prior probabilities $P(E_1)$ and $P(E_2)$.

Q: Wait, why do we use the total probability $P(S)$ in the denominator?

Think of $P(S)$ as your 'new universe.' Since we already know the motorcycle is standard, all non-standard motorcycles are removed from consideration. The denominator $P(S)$ acts as a normalization factor, ensuring that the sum of probabilities for all possible sources (Plant A and Plant B) equals 1 within this restricted universe.

Q: Is there a shortcut to avoid decimals?

Absolutely! You can work with fractions or even counts. Imagine a batch of 100 motorcycles. 60 come from A (48 standard) and 40 from B (36 standard). Total standard = 84. The probability from B is simply 36/84. Working with integers often prevents silly calculation errors during the exam.

Q: What if the question asked for the probability of it being from Plant A instead?

You would simply calculate $P(E_1|S) = \frac{P(E_1)P(S|E_1)}{P(S)}$. Notice that $P(E_1|S) + P(E_2|S)$ must equal 1. If you find one, you can always find the other by subtracting from unity!

Q: Why does the question ask for $126p$ instead of just $p$?

In JEE Advanced, this is a common technique to ensure the final answer is an integer. It prevents students from guessing based on decimal formats and forces you to carry the fraction through to the very end, testing your precision and algebraic discipline.

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

Options:

(A) 54

(B) 66

(C) 64

(D) 56

Sigma Percentile · Accepted Answer

### The Detective Work of Probability: Unraveling the Source

Imagine you are a quality control inspector at a massive motorcycle manufacturing conglomerate. You walk into the warehouse, pick a motorcycle at random, and find it is of 'Standard Quality.'

Your boss asks you, 'What is the probability this came from Plant B?' This isn't just a math problem; it is a classic detective story. We have the evidence (the motorcycle is standard), and we need to determine the culprit (the plant of origin).

Welcome to the elegant world of **Bayes' Theorem**.

### Phase 1: Mapping the Territory

Before we dive into the numbers, we must define our events clearly. Let $E_1$ be the event that our motorcycle was born in Plant A, and $E_2$ be the event it came from Plant B.

We know the production distribution: $P(E_1) = 0.6$ and $P(E_2) = 0.4$. These are our 'prior' probabilities—what we know before we even look at the quality of the bike.

Next, we define the quality event $S$. We are told that Plant A has an 80% success rate, so $P(S|E_1) = 0.8$.

Plant B is even more efficient, with a 90% success rate, so $P(S|E_2) = 0.9$. These are our conditional probabilities, which tell us how likely the evidence ($S$) is, given a specific cause ($E_1$ or $E_2$).

### Phase 2: The Total Probability Landscape

To find the probability that our motorcycle came from Plant B, we first need to understand the 'Total Probability' of picking a standard motorcycle. Think of this as the sum of all possible ways to get a standard bike.

Mathematically, we express this as:

$$P(S) = P(E_1) \cdot P(S|E_1) + P(E_2) \cdot P(S|E_2)$$

When we plug in our values, we get $P(S) = (0.6 	imes 0.8) + (0.4 	imes 0.9)$. This simplifies to $0.48 + 0.36 = 0.84$.

This $0.84$ represents the total 'pool' of standard motorcycles in our warehouse. It is the denominator of our final answer, the normalization constant that makes our probability space complete.

### Phase 3: The Bayes' Theorem Breakthrough

Now, we apply the magic of Bayes' Theorem. We want the probability of $E_2$ given $S$. The formula is:

$$p = P(E_2|S) = \frac{P(E_2) \cdot P(S|E_2)}{P(S)}$$

The numerator, $P(E_2) \cdot P(S|E_2) = 0.36$, represents the specific contribution of Plant B to the total pool of standard motorcycles. By dividing this by the total probability $P(S) = 0.84$, we are essentially asking: 'Out of all the standard motorcycles, what fraction came from Plant B?'

Calculating this gives us $p = \frac{0.36}{0.84}$. By multiplying both numerator and denominator by 100, we get $\frac{36}{84}$.

Simplifying this fraction by dividing by 12, we arrive at the beautiful, clean result: $p = \frac{3}{7}$.

### Phase 4: The Final Victory

The problem asks us for $126p$. Now that we have our value for $p$, the final step is a simple arithmetic flourish:

$$126p = 126 	imes \frac{3}{7}$$

Dividing 126 by 7 gives us 18, and $18 	imes 3$ brings us to our final destination: **54**.

Look at what you have achieved! You navigated through the production percentages, accounted for the conditional quality rates, applied the inverse probability theorem, and simplified the final expression.

This is the essence of JEE Advanced mathematics—taking a complex, real-world scenario and distilling it into a precise, logical truth. Keep this clarity of thought, and no problem will ever be too daunting.

Select Answer:

Visualized Solution

Defining the Production Events $E_{1}$ and $E_{2}$

Defining the Quality Event $S$

Identifying the Goal with Bayes' Theorem

Total Probability of Standard Quality $P (S)$

Calculating Total Probability $P (S)$

Applying Bayes' Theorem for $p$

Calculating the Value of $p$

Final Calculation for $126 p$

The Sigma Insight: Bayes' Theorem

The Detective Work of Probability

Phase 1

Phase 2

Phase 3

Phase 4

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Bag A contains 2 white, 1 black and 3 red balls and bag B contains 3 black, 2 red and $n$ white balls. One bag is chosen at random and 2 balls drawn from it at random, are found to be 1 red and 1 black. If the probability that both balls come from Bag A is $\frac{6}{11}$ , then $n$ is equal to ____.