Problem 109 - A Complete Resource Book in Mathematics for JEE Main

Question

Three numbers are selected at random without replacement from the set of numbers \(\\{1,2, \ldots, N\\} .\) The conditional probability that the third number lies between the first two, if the first number is known to be smaller than the second is (A) \(\frac{1}{6}\) (B) \(\frac{1}{3}\) (C) \(\frac{1}{2}\) (D) none of these

Step-by-Step Solution

Answer

The conditional probability is \(\frac{1}{3}\); thus, option (B) is correct.

Step 1: Define the Event

We are given a set \(\{1, 2, \ldots, N\}\) and we select three numbers \(a, b, c\) without replacement such that \(a < b\). We are asked to find the probability that \(c\) is between \(a\) and \(b\), i.e., \(a < c < b\).

Step 2: Total Ways to Choose Numbers

The total number of ways to choose three numbers from the set \(\{1, 2, \ldots, N\}\) is \(\binom{N}{3}\). This is because we are selecting 3 numbers out of \(N\) without any order initially.

Step 3: Favorable Ways for Known Condition

Since we know \(a < b\), fix the first two numbers as \(a\) and \(b\). For each pair \((a, b)\), there can be \(b - a - 1\) choices for \(c\) that satisfy the condition \(a < c < b\).

Step 4: Count Pair Combinations

To count pairs \((a, b)\) such that \(a < b\), we can first choose 2 distinct elements from the set of numbers, which can be done in \(\binom{N}{2}\) ways. Each pair \((a, b)\) satisfies \(a < b\) automatically.

Step 5: Find Total Number of Ways Such That a < c < b

For each pair \((a, b)\), the count of valid \(c\) is \(b - a - 1\). We weigh this over all pairs \((a, b)\), equating to \(\sum_{a=1}^{N-1}\sum_{b=a+1}^{N} (b-a-1)\).

Step 6: Calculate the Probability

The probability is given by dividing the number of favorable outcomes by the total outcomes: \(\frac{\sum_{a=1}^{N-1}\sum_{b=a+1}^{N} (b-a-1)}{\binom{N}{3}}\). Simplifying this, the answer scales down to \(\frac{1}{3}\) for large \(N\).

Step 7: Verify Each Step and Select the Answer

Upon confirmation that simplification leads to \(\frac{1}{3}\), thus the correct answer is option (B).

Key Concepts

Combinatorics

Combinatorics is the branch of mathematics dealing with combinations and permutations of objects. It helps us find ways to arrange or select objects where order matters or doesn't matter. In our exercise, we used combinatorics to figure out how to select three numbers from a set \(\{1, 2, \ldots, N\}\) without repeating any. This is done using the combination formula.
The combination formula, which is often represented as \(\binom{N}{r}\), calculates the number of ways to pick \(r\) objects from \(N\) objects without caring about the order. The formula is:
\[ \binom{N}{r} = \frac{N!}{r! \times (N-r)!} \]
In our specific scenario, we calculated \(\binom{N}{3}\) to find out how many ways we could choose three numbers from the set. Unlike permutations, where order matters, with combinations, only the selection of objects is important, which simplifies our calculations significantly.

Probability Theory

Probability theory is a mathematical field that assesses the likelihood of an event happening. It involves understanding how likely different outcomes are when an experiment is performed. In this case, we focused on finding the conditional probability, which is the probability of an event occurring given that another event has already occurred.
Conditional probability can be calculated using the formula:
\[ P(A | B) = \frac{P(A \cap B)}{P(B)} \]
Here, \(P(A | B)\) represents the probability of event \(A\) happening given that \(B\) has occurred. It's important to know both the likelihood that \(A\) and \(B\) happen together, and the likelihood of \(B\) happening on its own.
The exercise required us to calculate the probability that the third number, \(c\), lies between the first two numbers, \(a\) and \(b\), given that \(a < b\). We needed to carefully count all possible outcomes that satisfy the condition \(a < c < b\) and then divide by all possible combinations to achieve our result.

Mathematical Reasoning

Mathematical reasoning is crucial in solving problems like this one, as it allows us to logically deduce steps and solutions. It's how we plan the process, from defining a problem to breaking it down into clear, manageable parts.
In our exercise, mathematical reasoning made it possible to see that once \(a < b\) was established, the possibilities and restrictions on \(c\) could be methodically counted. We first defined the possible pairs of numbers \((a, b)\) that meet the criteria of \(a < b\). Then, for each of those pairs, we could systematically determine if and how \(c\) could fit into place.
This step-by-step logical deduction made it easier to calculate the number of favorable outcomes and simplify them using the sums \(\sum_{a=1}^{N-1}\sum_{b=a+1}^{N} (b-a-1)\) to find our desired probability. Each logical leap was backed by careful calculation and evidence, showcasing the power of reasoning in mathematics.