Sigma Percentile
JEE Advanced 2015
LEVELJEE Advanced

Animated Solution for Mathematics - Probability: Comprehension Passage

Let and be the number of red and black balls, respectively, in box I. Let and be the number of red and black balls, respectively, in box II.
Question 1:

One of the two boxes, box I and box II, was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box II is 1/3, then the correct option(s) with the possible values of and is(are)

Select Answer:

* Multiple Correct
Question 2:

A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from box I, after this transfer, is 1/3, then the correct option(s) with the possible values of and is(are)

Select Answer:

* Multiple Correct

Visualized Solution

Visualizing the Two Boxes

  • Let Box I contain red balls and black balls.
  • Let Box II contain red balls and black balls.
  • We will analyze two distinct probability experiments using these boxes.

Understanding the First Experiment

  • Experiment 1: A box is selected at random, and a ball is drawn.
  • Let be the event of choosing Box I, and be the event of choosing Box II.
  • Since the box is chosen at random: .
  • Let be the event that the drawn ball is Red.

Applying Bayes' Theorem

  • We are given that the drawn ball is Red ().
  • We need the conditional probability that it came from Box II: .
  • By Bayes' Theorem:
  • P(E_2 | R) = \frac{P(R | E_2) P(E_2)}{P(R | E_1) P(E_1) + P(R | E_2) P(E_2)}

Substituting the Probabilities

  • Probability of drawing Red from Box I:
  • Probability of drawing Red from Box II:
  • Substitute these into Bayes' Theorem:
  • \frac{1}{3} = \frac{\left(\frac{n_3}{n_3 + n_4}\right) \cdot \frac{1}{2}}{\left(\frac{n_1}{n_1 + n_2}\right) \cdot \frac{1}{2} + \left(\frac{n_3}{n_3 + n_4}\right) \cdot \frac{1}{2}}

Simplifying the Relation

  • Cancel the common factor of from numerator and denominator:
  • \frac{1}{3} = \frac{\frac{n_3}{n_3 + n_4}}{\frac{n_1}{n_1 + n_2} + \frac{n_3}{n_3 + n_4}}
  • Cross-multiply to simplify:
  • \frac{n_1}{n_1 + n_2} + \frac{n_3}{n_3 + n_4} = 3 \left(\frac{n_3}{n_3 + n_4}\right)
  • Subtract from both sides:
  • \frac{n_1}{n_1 + n_2} = 2 \left(\frac{n_3}{n_3 + n_4}\right)

Testing the Options for Question 1

  • We need to check which options satisfy:
  • Option (a): (Correct)
  • Option (b): (Correct)
  • Option (c): and Option (d): (Incorrect)

Setting Up the Transfer Experiment

  • Experiment 2: A ball is drawn from Box I and transferred to Box II.
  • Then, a ball is drawn from Box I.
  • We are given that the probability of drawing a Red ball from Box I after this transfer is .

Total Probability Theorem for Transfer

  • The transferred ball can be either Red () or Black ().
  • Probability of transferring Red:
  • Probability of transferring Black:
  • By the Law of Total Probability:
  • P(R_{\text{after}}) = P(R | T_R) P(T_R) + P(R | T_B) P(T_B)$

Substituting the Conditional Probabilities

  • If Red is transferred (): Box I now has Red and Black balls.
  • \implies P(R | T_R) = \frac{n_1 - 1}{n_1 + n_2 - 1}
  • If Black is transferred (): Box I now has Red and Black balls.
  • \implies P(R | T_B) = \frac{n_1}{n_1 + n_2 - 1}
  • Substitute into the total probability formula:
  • P(R_{\text{after}}) = \left(\frac{n_1 - 1}{n_1 + n_2 - 1}\right) \left(\frac{n_1}{n_1 + n_2}\right) + \left(\frac{n_1}{n_1 + n_2 - 1}\right) \left(\frac{n_2}{n_1 + n_2}\right)

The Elegant Simplification

  • Factor out the common terms from the numerator:
  • P(R_{\text{after}}) = \frac{n_1 (n_1 - 1) + n_1 n_2}{(n_1 + n_2 - 1)(n_1 + n_2)}
  • Simplify the numerator:
  • Cancel the common term :
  • P(R_{\text{after}}) = \frac{n_1 (n_1 + n_2 - 1)}{(n_1 + n_2 - 1)(n_1 + n_2)} = \frac{n_1}{n_1 + n_2}

Verifying Options for Question 2

  • We are given:
  • Let's test the options:
  • Option (a):
  • Option (b):
  • Option (c): (Correct)
  • Option (d): (Correct)

The Sigma Insight: Bayes' Theorem

Solution Diagram

The Dance of Uncertainty

Mastering Probability
Welcome, future engineer. Today, we are not just solving a probability problem; we are peeling back the layers of uncertainty to reveal the elegant symmetry hidden within.
Probability is the language of the universe, and in problems like this, it often feels like we are walking through a maze. But fear not—we have the map.

Phase 1

The Bayes' Theorem Trap
Imagine you are standing before two boxes, Box I and Box II. You do not know which one you are holding, but you know the odds of picking either are equal: .
You draw a ball, and it is red. Suddenly, the world shifts. You are asked: "What is the probability this came from Box II?"
This is the classic "inverse" problem. We are not asking "What is the chance of drawing red?", we are asking "Given the red ball, what is the chance of the source?"
This is where Bayes' Theorem shines. It is the bridge between the effect and the cause. We write it as:
When we plug in our values, notice the magic. The factor of appears in every term and cancels out.
We are left with a beautiful, clean relationship: the ratio of red balls in Box I must be exactly twice the ratio of red balls in Box II. This is the geometric reality of your setup.
If you ever feel lost, return to this ratio:

Phase 2

The Transfer Paradox
Now, let us introduce a dynamic element. We transfer a ball from Box I to Box II. It feels like the system is becoming chaotic, as we do not know if we transferred a red ball or a black ball.
Does this ruin our probability? Not at all. This is the Law of Total Probability in action.
We consider two mutually exclusive universes: one where we transferred a red ball () and one where we transferred a black ball (). The probability of drawing a red ball after the transfer is the weighted sum of these two universes:
I know this algebraic expression looks terrifying, especially when you see fractions like . Take a breath.
When you expand this, something miraculous happens. The terms rearrange, and the denominator cancels out with the numerator.
You are left with the original probability: .

The Lesson

Why does this happen? It is the beauty of invariance. The transfer process, while changing the physical count of balls, preserves the expected probability of drawing a red ball.
It is a reminder that in physics and mathematics, some quantities remain conserved even when the system appears to change. Keep this intuition close to your heart.
When you face complex problems in the exam hall, look for the symmetry, look for the cancellation, and trust the math. You have the tools; now go forth and conquer.

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