Problem 18 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Two integers are selected at random from the set \(\\{1,2, \ldots, 11\\}\). Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is : \([\) Jan. \(11,2019(I)]\) (a) \(\frac{7}{10}\) (b) \(\frac{1}{2}\) (c) \(\frac{2}{5}\) (d) \(\frac{3}{5}\)

Step-by-Step Solution

Answer

The probability is \(\frac{2}{5}\), corresponding to option (c).

Step 1: Identify Even and Odd Numbers

The set is \(\{1, 2, \ldots, 11\}\). Even numbers are \(\{2, 4, 6, 8, 10\}\) and odd numbers are \(\{1, 3, 5, 7, 9, 11\}\). There are 5 even numbers and 6 odd numbers.

Step 2: Understand Conditions

The condition is that the sum of the two selected numbers is even. If both numbers are even or both are odd, their sum will be even.

Step 3: Calculate Favorable Outcomes

Case 1: Both numbers are even. There are 5 even numbers, so the number of ways to select 2 even numbers is \( \binom{5}{2} = 10 \). Case 2: Both numbers are odd. There are 6 odd numbers, so the number of ways to select 2 odd numbers is \( \binom{6}{2} = 15 \).

Step 4: Total Favorable Outcomes

Add the outcomes from both cases: \(10\) (both even) + \(15\) (both odd) = \(25\) ways.

Step 5: Calculate Probability of Both Even

The probability that both numbers are even given that the sum is even is calculated using the formula: \( \frac{\text{Favorable outcomes for both even}}{\text{Total outcomes where sum is even}} = \frac{10}{25} = \frac{2}{5} \).

Key Concepts

Even Numbers

In mathematics, an even number is any integer that can be divided by 2 without leaving a remainder. A typical way to express an even number is as \( 2n \), where \( n \) is an integer. This characteristic makes even numbers quite straightforward to identify: they end with 0, 2, 4, 6, or 8. When considering combinatorics problems like selecting numbers from a set, knowing that even numbers add certain properties, like symmetry in addition, can simplify solving puzzles. For instance, the sum of two even numbers is always even because \( 2a + 2b = 2(a+b) \). This property is key in problems involving conditional probability where the condition depends on the parity (even or odd nature) of numbers.

Odd Numbers

Odd numbers are integers that are not evenly divided by 2. These numbers can be expressed as \( 2n + 1 \), where \( n \) is an integer. Odd numbers have a distinct set of properties, making them interesting in problem-solving. For example, the sum of two odd numbers results in an even number \((2m + 1) + (2n + 1) = 2(m+n+1)\), a fact frequently used in probability exercises. Understanding such properties is very helpful when working with questions concerning the sums of randomly selected integers, as these properties can simplify identifying possible outcomes that satisfy specific conditions.

Combinatorics

Combinatorics is the branch of mathematics concerning counting, arrangement, and combination. It delves into topics like permutations and combinations, which help solve problems like figuring out the number of ways to select subsets from sets.
When dealing with even and odd numbers from a defined set, such as \( \{1, 2, \, \ldots, 11\} \), combinatorial concepts are vital. To determine possible combinations that meet specific criteria, like both numbers selected being even, we use combinations symbolically represented by \( \binom{n}{k} \). This notation counts the number of ways to choose \( k \) outcomes from \( n \) possibilities. In our context, this method helps us calculate the number of ways to choose 2 numbers that meet a given condition like both being even.

Probability

Probability is a mathematical study of random events. It measures the likelihood of a specific event occurring. In conditional probability, the focus shifts to reassessing likelihood given a certain condition.
For example, in this exercise, we calculate the probability that both selected numbers are even, conditioned by the total sum being even. This involves recognizing valid combinations that match the condition and calculating their ratio against all favorable combinations.
The formula employed here is fundamental:

• \( P(A|B) = \frac{P(A \cap B)}{P(B)} \)

In simpler terms, it's the fraction where the numerator counts desired outcomes (both numbers even) and the denominator counts all possible favorable outcomes (sum is even). Understanding how to break down problems into manageable parts using this approach is crucial for mastering conditional probability.