Problem 22 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

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: [Jan. 09, 2019 (II)] (a) \(\frac{21}{49}\) (b) \(\frac{27}{49}\) (c) \(\frac{26}{49}\) (d) \(\frac{32}{49}\)

Step-by-Step Solution

Answer

\( \frac{32}{49} \)

Step 1: Define Initial Probabilities

An urn initially contains 5 red and 2 green balls, totaling 7 balls. The probability of drawing a red ball first is \( \frac{5}{7} \) and the probability of drawing a green ball first is \( \frac{2}{7} \).

Step 2: Calculate Outcome if First Ball is Green

If the first ball is green, one green ball is removed and one red ball added, resulting in 6 red balls and 1 green ball left in the urn. Thus, after this draw, the probability of drawing a red ball next is \( \frac{6}{7} \).

Step 3: Calculate Outcome if First Ball is Red

If the first ball is red, one red ball is removed and one green ball added, resulting in 4 red balls and 3 green balls. Thus, after this draw, the probability of drawing a red ball next is \( \frac{4}{7} \).

Step 4: Use Total Probability Theorem

The probability of the second ball being red can be computed using the total probability theorem: \( P(Red_2) = P(Red_2 | Green_1) \cdot P(Green_1) + P(Red_2 | Red_1) \cdot P(Red_1) \).

Step 5: Substitute Values and Compute

Substitute known values into the total probability formula: \[P(Red_2) = \left(\frac{6}{7}\right) \cdot \left(\frac{2}{7}\right) + \left(\frac{4}{7}\right) \cdot \left(\frac{5}{7}\right)\]Calculate this: \[P(Red_2) = \frac{12}{49} + \frac{20}{49} = \frac{32}{49}\].

Step 6: Conclusion

The probability that the second ball drawn is red is \( \frac{32}{49} \). Thus, the correct answer is (d).

Key Concepts

Conditional Probability

Conditional probability helps us determine the likelihood of an event occurring given that another event has already happened. In the context of our urn problem, we are interested in finding the probability of drawing a red ball as the second ball under different conditions based on the first ball's color.

When considering conditional probability, you can think of it as a way to refine probabilities by incorporating additional information.
For example:

• If the first ball drawn is green, what is the probability the next will be red? Here, we consider the condition of the first draw.
• Similarly, what's the probability the second ball will be red if the first was red?

These are situations where conditional probability shines, allowing us to make more accurate predictions based on given conditions.

Total Probability Theorem

The total probability theorem is a powerful tool in probability theory that helps us compute an overall probability by breaking it down into mutually exclusive events.

In the urn problem, the total probability theorem allows us to account for all possible ways the second ball can be red. We first calculate the probability for each possible first event (first ball green or red), and then multiply these by the probability of the second ball being red, given each first event:

• Probability of second ball being red given the first was green: \( P(Red_2 | Green_1) \)
• Probability of second ball being red given the first was red: \( P(Red_2 | Red_1) \)

The final probability is then found by summing these weighted probabilities, considering the probability of each starting condition, using the formula:\[ P(Red_2) = P(Red_2 | Green_1) \cdot P(Green_1) + P(Red_2 | Red_1) \cdot P(Red_1) \]

Discrete Mathematics

Discrete mathematics deals with structures that are fundamentally distinct and separate. It involves things that can be counted in integers, like how many ways you can draw balls from the urn. It provides the tools for reasoning through processes that don’t involve continuity, instead focusing on finite or countable processes.

In the urn problem, we engage discrete mathematics when counting the number of red and green balls left after each draw and considering possible outcomes for the second draw. The probabilities calculated throughout depend on these counts and combinations. Here’s how discrete mathematics applies:

• The total number of balls is a fixed value, impacting the probability calculations.
• Each event (drawing a specific colored ball) is a distinct occurrence that can be counted.
• The transitions (adding balls of a specific color) also follow discrete steps.

By using principles from discrete mathematics, we can manage and solve problems involving a set structure and sequence, just like our urn experiment.