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

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

(B) \( \frac{3(r-1)(r-2)}{(m+1)(m+2)(m+3)} \)

Step 1: Understand the Problem

We need to find the probability that all black balls (3 in total) have been drawn in exactly the \( r \)-th draw from a bag containing \( m \) white and 3 black balls, drawn without replacement. This means the 3rd black ball must be drawn on the \( r \)-th draw, and the other 2 black balls have to be drawn in the earlier \( r-1 \) draws.

Step 2: Select 2 Black Balls in the First \( r-1 \) Draws

Calculate the ways to draw 2 black balls in the first \( r-1 \) draws from the total of 3 black balls. This can be chosen in \( \binom{r-1}{2} \) ways.

Step 3: Arrange \( r-3 \) White Balls Among the First \( r-1 \) Draws

The remaining \( r-3 \) draws (since we drew 2 black balls already in step 2) are white balls. There are \( m \) white balls initially, and we are drawing and arranging \( r-3 \) of them.

Step 4: Draw the Final Black Ball on the \( r \)-th Draw

The \( r \)-th draw must be the 3rd black ball. Hence, the \( r \)-th draw has only one outcome: drawing the last black ball. This occurs with a probability of 1, due to the setup.

Step 5: Calculate Total Ways to Draw with Replacement

Calculate the total number of ways to arrange all draws. We fill the first \( r \) draws with \( r \) balls chosen from \( m+3 \). The number of possible draw sequences totals \( \binom{m+3}{r} \).

Step 6: Calculate the Probability

The probability needed is computed from step 2 to step 5: \( \Pr(\text{procedure ends at } r\text{-th draw}) = \frac{\binom{r-1}{2} \cdot \text{arrangement of white balls}}{\binom{m+3}{r}} \). Simplify to obtain: \( \Pr = \frac{3(r-1)(r-2)}{(m+1)(m+2)(m+3)} \).

Step 7: Conclusion

Comparing our calculated result with the options given in the problem statement, we finalize: The probability that the procedure ends at the \( r \)-th draw is answer (B): \( \frac{3(r-1)(r-2)}{(m+1)(m+2)(m+3)} \).

Key Concepts

Binomial coefficients

Binomial coefficients are fundamental in various fields of mathematics, especially in combinatorics. They represent the number of ways to choose a subset of elements from a larger set, without regard to the order of selection.

Mathematically, the binomial coefficient is denoted as \( \binom{n}{k} \), which reads as 'n choose k'. This expression computes the number of ways to choose \( k \) elements from a larger set of \( n \) elements. The formula for calculating it is given by: \[\binom{n}{k} = \frac{n!}{k!(n-k)!}\] Here, \(!\) denotes factorial, which is the product of all positive integers up to a specified number.

In the context of the ball drawing problem, binomial coefficients help us determine the different ways to draw black balls from a set. For instance, if we have 3 black balls and need to draw 2 of them within the first \( r-1 \) draws, we use the binomial coefficient \( \binom{r-1}{2} \) to count these possible selections.

Combinatorial probability

Combinatorial probability involves using counting methods to determine the likelihood of a particular event. It is particularly useful when dealing with discrete random events, such as drawing balls from a bag. In this context, probability is calculated by comparing favorable outcomes to the total number of possible outcomes.

For the ball drawing problem, we are interested in the probability that the sequence of draws ends exactly at the \( r \)-th draw, with all black balls being drawn. To find this probability, we calculate the number of ways to achieve this precise sequence and divide it by the total possible sequences of draws.

To determine these counts, we utilize binomial coefficients to calculate the possible configurations of drawing specific balls. This approach exemplifies how combinatorial methods provide a systematic way to address probability challenges by breaking down the problem into countable outcomes.

Ball drawing problem

The ball drawing problem is a classic example used to illustrate concepts in probability and combinatorics. It involves drawing balls from a bag until a certain condition is met. In our specific exercise, we aim to find the probability that all black balls are drawn by the \( r \)-th draw from a mixture of black and white balls without replacement.

This requires understanding the rules of the problem:

• Select balls one by one until all black balls are drawn.
• The finishing line, so to speak, is the \( r \)-th draw. The challenge is ensuring that the third black ball is drawn exactly at this moment.

The key steps in solving include using binomial coefficients for choosing the black balls in the former draws, filling the remainder with white balls, and ensuring the final draw is the third black ball. The calculated probability is then contrasted with the total possible outcomes of drawing \( r \) balls, leading us to the correct probability expression and the solution to the given options.