JEE Main 2019 (9 January)

LEVELJEE Main

Animated Solution for Mathematics - Probability: An urn contains 5 red and 2 green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn. Now, a second ball is drawn at random from it. The probability that the second ball is red, is :

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the urn as a dynamic system. The state of the urn for the second draw is not fixed; it is a direct consequence of the first draw. You must treat the two possible outcomes of the first draw as separate, mutually exclusive branches of reality.

Hint 2

Hint 2 (The Arsenal): To find the probability of the second event, use the Law of Total Probability. This theorem allows you to calculate the probability of an event by summing the probabilities of all possible ways that event can occur, weighted by the probability of each path.

Hint 3

Hint 3 (The Execution): Start by defining your events:

R_{1}

(Red first),

G_{1}

(Green first), and

R_{2}

(Red second). Your master equation is

P (R_{2}) = P (R_{1}) \times P (R_{2} ∣ R_{1}) + P (G_{1}) \times P (R_{2} ∣ G_{1})

. Calculate each component carefully based on the updated urn composition.

Visualized Solution

Visualizing the Initial State

Initial contents: 5 Red (R) and 2 Green (G) balls.
Total balls: $5 + 2 = 7$ .

Understanding the Experiment Rules

Draw a ball at random.
If Green, add a Red ball.
If Red, add a Green ball.
Original ball is not returned.

The Law of Total Probability

Let $R_{1}, G_{1}$ be the events for the first draw.
Let $R_{2}$ be the event that the second ball is Red.
$P (R_{2}) = P (R_{1}) \cdot P (R_{2} ∣ R_{1}) + P (G_{1}) \cdot P (R_{2} ∣ G_{1})$

Case 1: First Ball is Red

Probability of drawing Red first: $P (R_{1}) = \frac{5}{7}$
Action: Remove 1 Red, Add 1 Green.

State After Drawing Red

New State: $(5 - 1) R, (2 + 1) G = 4 R, 3 G$
Total balls remain $7$ .
Conditional Probability: $P (R_{2} ∣ R_{1}) = \frac{4}{7}$

Case 2: First Ball is Green

Probability of drawing Green first: $P (G_{1}) = \frac{2}{7}$
Action: Remove 1 Green, Add 1 Red.

State After Drawing Green

New State: $(5 + 1) R, (2 - 1) G = 6 R, 1 G$
Total balls remain $7$ .
Conditional Probability: $P (R_{2} ∣ G_{1}) = \frac{6}{7}$

Setting Up the Equation

Substitute values into the Total Probability formula:
$P (R_{2}) = (\frac{5}{7} \times \frac{4}{7}) + (\frac{2}{7} \times \frac{6}{7})$

Executing the Multiplication

Calculate the first term: $\frac{5}{7} \times \frac{4}{7} = \frac{20}{49}$
Calculate the second term: $\frac{2}{7} \times \frac{6}{7} = \frac{12}{49}$
$P (R_{2}) = \frac{20}{49} + \frac{12}{49}$

Final Addition

Since denominators are the same, add the numerators:
$P (R_{2}) = \frac{20 + 12}{49}$
$P (R_{2}) = \frac{32}{49}$

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The Sigma Insight: Total Probability Theorem

Solution Diagram

Analyzing the Setup

The experiment begins with an urn containing

5

red balls and

2

green balls. The total number of balls is

5 + 2 = 7

The system follows a specific rule: when a ball is drawn, it is not replaced. Instead, if a green ball is drawn, a red ball is added; if a red ball is drawn, a green ball is added. Consequently, the total number of balls in the urn remains constant at

7

for the second draw.

The Master Equation

To find the probability that the second ball is red, denoted as

P (R_{2})

, we utilize the Law of Total Probability. This event occurs through two mutually exclusive scenarios: drawing a red ball first (

R_{1}

) or drawing a green ball first (

G_{1}

The governing equation is:

P (R_{2}) = P (R_{1}) \times P (R_{2} ∣ R_{1}) + P (G_{1}) \times P (R_{2} ∣ G_{1})

Path 1

The Red Branch

If the first ball drawn is red, the probability is

P (R_{1}) = \frac{5}{7}

. Following the rules, we remove one red ball and add one green ball.

The urn composition shifts from

5 R, 2 G

4 R, 3 G

. In this state, the conditional probability of drawing a red ball is

P (R_{2} ∣ R_{1}) = \frac{4}{7}

The probability for this branch is:

P (R_{1}) \times P (R_{2} ∣ R_{1}) = \frac{5}{7} \times \frac{4}{7} = \frac{20}{49}

Path 2

The Green Branch

If the first ball drawn is green, the probability is

P (G_{1}) = \frac{2}{7}

. Following the rules, we remove the green ball and add a red one.

The urn composition shifts from

5 R, 2 G

6 R, 1 G

. In this state, the conditional probability of drawing a red ball is

P (R_{2} ∣ G_{1}) = \frac{6}{7}

The probability for this branch is:

P (G_{1}) \times P (R_{2} ∣ G_{1}) = \frac{2}{7} \times \frac{6}{7} = \frac{12}{49}

Final Calculation

To determine the total probability, we sum the results of the two independent paths:

P (R_{2}) = \frac{20}{49} + \frac{12}{49}

Adding the numerators, we arrive at the final result:

$P (R_{2}) = \frac{32}{49}$

Common Traps & FAQs

Wait, why do we use the Law of Total Probability? Can't we just count the total number of outcomes?

That is a great question! While counting works for simple, static scenarios, this problem is dynamic. The composition of the urn changes based on the first draw. The Law of Total Probability is the perfect tool here because it forces us to account for the 'history' of the urn. It ensures we don't accidentally mix up the two distinct paths (Red-first vs. Green-first) that lead to the second draw.

Why does the total number of balls remain 7? I thought we removed one?

Excellent observation! You are right that we remove the first ball. However, the rules state that if we draw a green ball, we add a red one, and vice versa. So, we remove one ball and add one ball. The net change to the total count is zero.

7 - 1 + 1 = 7

. Always keep a close eye on the 'net' change in these types of problems!

What if the question asked for the probability of the second ball being green? How would the logic change?

The beauty of the Law of Total Probability is its flexibility! If you wanted

P (G_{2})

, you would simply change your target event. The formula would become

P (G_{2}) = P (R_{1}) \times P (G_{2} ∣ R_{1}) + P (G_{1}) \times P (G_{2} ∣ G_{1})

. The structure remains identical; only the conditional probabilities change based on the new urn states.