Problem 99 - A Complete Resource Book in Mathematics for JEE Main

Question

Three natural numbers are taken at random from the set \(A=\\{x: 1 \leq x \leq 100, x \in N\\}\). The probability that the \(A . M\). of the numbers taken is 25 is (A) \(\frac{{ }^{77} C_{2}}{{ }^{100} C_{3}}\) (B) \(\frac{{ }^{25} C_{2}}{{ }^{25} C_{3}}\) (C) \(\frac{{ }^{74} C_{2}}{{ }^{100} C_{3}}\) (D) none of these

Step-by-Step Solution

Answer

The probability is \(\frac{^{74}C_2}{^{100}C_3}\). Option (C).

Step 1: Define Arithmetic Mean Condition

The arithmetic mean (A.M.) of three numbers \(a, b, c\) is defined as \(\frac{a+b+c}{3} = 25\). This simplifies to \(a+b+c = 75\).

Step 2: Choose Numbers and Formulate Constraints

Consider the numbers \(a, b, c\) such that \(1 \leq a, b, c \leq 100\) and \(a+b+c=75\). These numbers need to be natural numbers, meaning \(a, b, c \geq 1\).

Step 3: Count Total Possible Selections of Three Numbers

The total ways to choose any three numbers from the set \(A\) of 100 elements is expressed as \(^{100}C_3\).

Step 4: Solve for Number of Ways under A.M. Condition

The problem of finding non-negative integer solutions to \(a+b+c = 75\) translates into a stars and bars combinatorial problem. We can translate the bounds into a constraint-solving problem involving distinct increments on each solution so \(a, b, c \geq 1\), implying \(a'=a-1, b'=b-1, c'=c-1\) and \(a'+b'+c' = 72\) with now \(a', b', c' \geq 0\).

Step 5: Apply Combinatorial Method to Count Valid Selections

Using the stars and bars method, the number of solutions to \(a'+b'+c' = 72\) is calculated as \(^{74}C_2\). This accounts for choosing 2 dividers in a sequence of 74 positions (72 stars and 2 bars).

Step 6: Calculate the Probability

The probability that the arithmetic mean of three chosen numbers is 25 is the ratio of desired outcomes to total outcomes: \(\frac{^{74}C_2}{^{100}C_3}\).

Key Concepts

Combinatorial Mathematics

Combinatorial mathematics is a branch of mathematics that deals with counting, arrangement, and combination of elements in sets. It involves studying the ways in which discrete structures can be arranged, combined, or selected from a larger set. In the context of our exercise, we are interested in the number of ways to select three natural numbers whose arithmetic mean is a specific value.To solve problems in combinatorics, we often use combinations, which are ways to select items from a group, without regard to their order. The formula for combinations is given by \( ^nC_k = \frac{n!}{k!(n-k)!} \), where \( n \) is the total number of items to choose from, and \( k \) is the number of items to select. For our problem, we use this formula to count distinct ways to select numbers and their required conditions.

Stars and Bars Method

The stars and bars method is a popular combinatorial technique used to solve problems related to distributing indistinguishable objects into distinct groups or bins. This method is particularly useful in situations where you have to find the number of ways to partition a whole into parts.In our problem, we used this method to find the number of ways to solve the equation \( a+b+c=75 \) with the constraint that \( a, b, c \geq 1 \). To apply stars and bars, we first transform the equation into a form that allows non-negative solutions. This is done by redefining our variables: set \( a' = a - 1 \), \( b' = b - 1 \), and \( c' = c - 1 \), thus transforming the equation into \( a' + b' + c' = 72 \). Here, the stars represent units we want to partition, and the bars denote dividers between partitions. The number of solutions is calculated by assigning positions for the bars amongst the stars, which is computed as \( ^{74}C_2 \) for the exercise.

Probability Calculation

Calculating probability involves determining the ratio of favorable outcomes to the total possible outcomes. Probability helps us quantify certainty in mathematical terms and is fundamental in decision-making and predictions.In our task, we are calculating the probability that the arithmetic mean of three randomly chosen natural numbers is 25, from numbers 1 to 100. Once we identified that there are \( ^{74}C_2 \) favorable ways to select numbers meeting the arithmetic mean condition, we needed to find the total combinations possible for choosing any three numbers from the set. We calculated this total using combinations \( ^{100}C_3 \), as there are 100 numbers to choose from and we are selecting 3. Thus, the probability is the fraction of favorable combinations over total combinations, given by \( \frac{^{74}C_2}{^{100}C_3} \).

Natural Numbers

Natural numbers are a fundamental set of numbers in mathematics starting typically from 1. They are the set of positive integers \( \{1, 2, 3, \ldots\} \), often used for counting and ordering. The concept of natural numbers is critical in combinatorics, probability, and various other mathematical domains.In this exercise, we were working with numbers within the natural set from 1 to 100. This set includes all whole numbers without fractions or negatives, making it simple to apply rules of combinatorics and arithmetic. The idea that the values \( a, b, c \) must be natural numbers (i.e., \( a, b, c \geq 1 \)) is essential for setting up our arithmetic mean and using combinatorial strategies like stars and bars, as it ensures all components are valid within their defined range.