Problem 37 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

There are \(n\) urns each containing \((n+1)\) balls such that the \(i^{\text {th }}\) urn contains \(i\) white balls and \((n+1-i)\) red balls. Let \(u_{i}\) be the event of selecting the \(i^{\text {th }}\) urn, \(i=1,2,3 \ldots, n\), and \(W\) denote the event of getting a white ball. (i) If \(P\left(u_{j}\right) \alpha i\), where \(i=1,2,3, \ldots, n\), then \(\lim _{n \rightarrow \infty} P(W)\) is equal to (a) 1 (b) \(1 / 4\) (c) \(2 / 3\) (d) \(3 / 4\)

Step-by-Step Solution

Answer

The correct option is (c) \(2/3\).

Step 1: Understanding the Problem

We have \(n\) urns, each containing \((n+1)\) balls where the \(i^{\text{th}}\) urn has \(i\) white balls and \((n+1-i)\) red balls. The event \(u_i\) refers to selecting the \(i^{\text{th}}\) urn, with probability \(P(u_i) \propto i\). Our goal is to find \(\lim_{n \to \infty} P(W)\), the probability of drawing a white ball.

Step 2: Probability of Drawing a White Ball from an Urn

If the \(i^{\text{th}}\) urn is selected, the probability of drawing a white ball from it is \(\frac{i}{n+1}\), since it contains \(i\) white balls out of \((n+1)\) balls.

Step 3: Calculating the Probability of Selecting Each Urn

Since \(P(u_i) \propto i\), we have \(P(u_i) = \frac{i}{\sum_{j=1}^{n} j} = \frac{i}{\frac{n(n+1)}{2}} = \frac{2i}{n(n+1)}\).

Step 4: Finding Overall Probability of Drawing a White Ball

The overall probability of drawing a white ball, \(P(W)\), is the sum of probabilities of drawing a white ball from each urn multiplied by the probability of selecting that urn: \[ P(W) = \sum_{i=1}^{n} P(W | u_i) \cdot P(u_i) = \sum_{i=1}^{n} \frac{i}{n+1} \cdot \frac{2i}{n(n+1)} = \frac{2}{(n+1)^2} \sum_{i=1}^{n} i^2 \]The sum \(\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}\).

Step 5: Simplifying the Expression for P(W)

Substitute \(\sum_{i=1}^{n} i^2\) into the equation for \(P(W)\):\[ P(W) = \frac{2}{(n+1)^2} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{n(2n+1)}{3(n+1)} \]

Step 6: Evaluating the Limit as n Approaches Infinity

Simplify the expression \(\frac{n(2n+1)}{3(n+1)}\) by dividing the numerator and the denominator by \(n\):\[ \lim_{n \to \infty} \frac{n(2n+1)}{3(n+1)} = \lim_{n \to \infty} \frac{2n + 1/n}{3 + 1/n} = \lim_{n \to \infty} \frac{2 + 1/n}{3 + 1/n} = \frac{2}{3} \]

Step 7: Conclusion

We have found that \(\lim_{n \to \infty} P(W) = \frac{2}{3}\). Therefore, the correct option is (c) \(2/3\).

Key Concepts

Urn problems

Urn problems are classic exercises in probability theory that help students understand complex concepts through a tangible metaphor. Imagine a series of urns, each containing balls of different colors. In the exercise, we have several urns, and each urn uniquely determines the composition of white and red balls. For example, the urn labeled with "1" has one white ball and the rest red, while the urn labeled "2" has two white balls, and so forth. This setup is key for understanding how probabilities are assigned and calculated. Key takeaways for solving urn problems involve:

• Understanding how each urn is filled and how its composition changes with each step.
• Knowing the probability of selecting any specific urn from the whole.
• Establishing how the composition of an urn influences the chance of drawing a particular color ball.

By grasping these property dynamics, students can set up probabilities and solve related questions using algebraic methods.

Infinite limits

Infinite limits in probability provide insight into long-term behavior as a parameter, often denoted by "n," approaches infinity. For instance, in the urn problem, we investigate what happens to the probability of drawing a white ball as the number of urns becomes very large.The core ideas here involve:

• Simplifying complex fractions into more manageable terms for manipulation.
• Identifying dominating variables as "n" increases to understand which elements become negligible.
• Applying limit laws that help solve equations as "n" approaches infinity.

As seen in the solution, the use of limits allows us to conclude that the overall probability converges to a definitive value, which in this case is \(\frac{2}{3}\). This is a crucial step to understanding how changing "n" shapes the probability outcome in infinite limit problems.

Algebraic manipulation

Algebraic manipulation is a cornerstone in solving probabilistic and mathematical problems. It entails using algebraic skills to rearrange, simplify, and solve equations. In the urn problem, algebraic manipulation was used in several steps to simplify the expressions, such as calculating the probability of selecting each urn and deriving the overall probability for drawing a white ball.Consider the key operations:

• Setting up correct equations to represent relationships among variables.
• Performing arithmetic and algebraic operations like factoring, expanding, and canceling terms appropriately.
• Using summation properties to simplify complex series, such as \(\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}\).

Applying algebraic manipulation not only leads to cleaner solutions but also makes it possible to draw conclusions about behavior as "n" approaches infinity, with outcomes like the limit evaluation yielding simple numbers like \(\frac{2}{3}\). Understanding these algebraic steps helps students tackle more complex problems with confidence.