Problem 31 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

A bag contains \(m\) white and 3 black balls. Balls are drawn one by one without replacement till all the black balls are drawn. The probability that this procedure for drawing balls will come to an end at the \(r\) th draw is (A) \(\frac{(r-1)(r-2)}{(m+1)(m+2)(m+3)}\) (B) \(\frac{3(r-1)(r-2)}{(m+1)(m+2)(m+3)}\) (C) \(\frac{2(r-1)(r-2)}{(m+1)(m+2)(m+3)}\) (D) none of these

Step-by-Step Solution

Answer

Option (B) is correct: \( \frac{3(r-1)(r-2)}{(m+1)(m+2)(m+3)} \).

Step 1: Understand the Problem

The problem requires us to find the probability that the process of drawing balls ends on the \(r\)th draw, where we have \(m\) white balls and 3 black balls. We need the last ball to be black, and we need all black balls drawn by the \(r\)th draw.

Step 2: Determine the Configuration

For the process to end at the \(r\)th draw, there must be exactly 2 black balls and \(r-3\) white balls drawn in the first \(r-1\) draws. The \(r\)th draw must be the last, i.e., the third black ball.

Step 3: Calculate Total Ways to Draw

The total number of ways to draw \(r\) balls from \(m+3\) balls is given by \( \binom{m+3}{r} \).

Step 4: Calculate Successful Events

The number of successful ways to draw the balls such that 2 black balls and \(r-3\) white balls are picked in the first \(r-1\) draws, followed by the last black ball at draw \(r\) is \( \binom{2}{2} \cdot \binom{m}{r-3} \), which simplifies to \( \binom{m}{r-3} \).

Step 5: Apply Probability Formula

We calculate the probability as the ratio of successful events to total possible events. This is \( \frac{\binom{m}{r-3}}{\binom{m+3}{r}} \).

Step 6: Compute the Binomial Coefficients

The binomial coefficient \( \binom{m}{r-3} = \frac{m!}{(r-3)!(m-r+3)!} \) and \( \binom{m+3}{r} = \frac{(m+3)!}{r!(m+3-r)!} \). Upon simplification, the ratio simplifies to \( \frac{3(r-1)(r-2)}{(m+1)(m+2)(m+3)} \).

Step 7: Identify the Correct Option

The probability matches with option (B): \( \frac{3(r-1)(r-2)}{(m+1)(m+2)(m+3)} \).

Key Concepts

Combinatorics

Combinatorics is an essential concept in mathematics that helps us understand how to arrange or combine different objects. In the context of probability exercises like the one in the original problem, combinatorics is used to calculate the number of ways certain outcomes can occur. For example, when you're drawing balls from a bag, combinatorics helps determine how many different combinations of black and white balls can be drawn without considering the order.

Consider the example where a certain number of balls are drawn from a bag. We use combinations when the order of selection does not matter. This uses the binomial coefficient, denoted as \( \binom{n}{k} \), which represents the number of ways to choose \( k \) elements from a set of \( n \) elements.

Combinatorics is powerful because it applies to so many scenarios beyond simple drawing exercises, such as seating arrangements, lottery probabilities, and even network connectivity measures in computer science. Understanding combinatorics is a stepping stone to mastering many complex probability and statistical challenges.

Binomial Coefficients

Binomial coefficients play a vital role in calculating combinations and are foundational to combinatorics. They are used when you want to find out in how many ways you can choose \( k \) objects from a set of \( n \) objects, which is essentially what happens when drawing from the bag of balls.

The binomial coefficient is represented as \( \binom{n}{k} \), pronounced "n choose k." It is calculated using the formula \[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]This formula accounts for all the possible ways to choose \( k \) objects, dividing by the order's redundancies (since order doesn't matter in combinations).

In the problem, the binomial coefficient helps compute the total number of ways and successful arrangements for drawing the balls. Understanding and applying binomial coefficients adequately can help simplify many problems, be they practical or theoretical, by breaking them down into more manageable parts.

Probability Without Replacement

Probability without replacement refers to the scenario where once an item is selected or drawn, it is not put back into the pool of possible selections. This affects the probability of subsequent events since each draw reduces the number of total possible outcomes.

In the given problem, as balls are drawn, they are not replaced. This means each draw alters the probability of drawing a black or white ball on the next attempt. The absence of replacement makes calculations slightly more complex because the sample space keeps changing.

For instance, when calculating the probability for the original problem to end on the \( r \)-th draw, considering the changing number of balls remaining is crucial. The scenario asks for exactly the third black ball on a specific draw, ensuring you've considered all previous draws where some balls are already removed.

A thorough understanding of probability without replacement is valuable in real-world applications, such as card games, quality control testing, and any situation where sampling is done consecutively without returning the sampled items.