Let the mean and variance of 7 observations 2, 4, 10, x, 12, 14, y, x > y, be 8 and 16 respectively. Two numbers are chosen from {1, 2, 3, x - 4, y, 5} one after another without replacement, then the probability, that the smaller number among the two chosen numbers is less than 4, is :
Q. Let the mean and variance of 7 observations 2, 4, 10, \(x\), 12, 14, \(y\), \(x > y\), be 8 and 16 respectively. Two numbers are chosen from \(\{1, 2, 3, x - 4, y, 5\}\) one after another without replacement, then the probability, that the smaller number among the two chosen numbers is less than 4, is :

(A) \(\frac{4}{5}\)

(B) \(\frac{3}{5}\)

(C) \(\frac{2}{5}\)

(D) \(\frac{1}{3}\)

Correct Answer: (A) \(\frac{4}{5}\)

Step-by-Step Detailed Solution

Step 1: Use Mean formula to find \(x + y\)
Mean (\(\bar{x}\)) = \(\frac{\sum x_i}{n} = 8\)
\(\frac{2 + 4 + 10 + x + 12 + 14 + y}{7} = 8\)
\(42 + x + y = 56\)
\(x + y = 14\) --- (Equation 1)

Step 2: Use Variance formula to find \(x^2 + y^2\)
Variance (\(\sigma^2\)) = \(\frac{\sum x_i^2}{n} - (\bar{x})^2 = 16\)
\(\frac{2^2 + 4^2 + 10^2 + x^2 + 12^2 + 14^2 + y^2}{7} - (8)^2 = 16\)
\(\frac{4 + 16 + 100 + x^2 + 144 + 196 + y^2}{7} - 64 = 16\)
\(\frac{460 + x^2 + y^2}{7} = 80\)
\(460 + x^2 + y^2 = 560\)
\(x^2 + y^2 = 100\) --- (Equation 2)

Step 3: Solve for \(x\) and \(y\)
We know \((x + y)^2 = x^2 + y^2 + 2xy\)
\((14)^2 = 100 + 2xy\)
\(196 = 100 + 2xy \Rightarrow 2xy = 96 \Rightarrow xy = 48\)
The numbers whose sum is 14 and product is 48 are 8 and 6.
Given \(x > y\), therefore \(x = 8\) and \(y = 6\).

Step 4: Identify the second set of numbers
The set is \(S = \{1, 2, 3, x-4, y, 5\}\)
Substituting \(x=8\) and \(y=6\):
\(S = \{1, 2, 3, 8-4, 6, 5\} = \{1, 2, 3, 4, 6, 5\}\)
Total elements \(n = 6\).

Step 5: Calculate Probability
Two numbers are chosen without replacement. Total ways = \(^6P_2 = 6 \times 5 = 30\).
We want the smaller number to be less than 4.
Let the two numbers be \((a, b)\). We want \(\min(a, b) < 4\).
This is easier to calculate using the complement: \(1 - P(\min(a, b) \ge 4)\).
The numbers \(\ge 4\) in the set are \(\{4, 5, 6\}\).
Ways to pick both numbers from this subset = \(^3P_2 = 3 \times 2 = 6\).
\(P(\min \ge 4) = \frac{6}{30} = \frac{1}{5}\).
Required Probability = \(1 - \frac{1}{5} = \frac{4}{5}\).

Related Theory: Statistics and Probability

This problem is a perfect blend of two fundamental pillars of the JEE Mathematics syllabus: Descriptive Statistics (Mean and Variance) and Classical Probability. It requires the ability to solve for unknowns and then apply combinatorial logic.

1. Measures of Dispersion: Variance

Variance measures how far a set of numbers is spread out from their average value. The most used formula in JEE is the computational form:

$$\sigma^2 = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2$$

In competitive exams, always use this form rather than \(\frac{\sum (x_i - \bar{x})^2}{n}\) as it simplifies algebraic manipulation when data points are integers.

2. Solving Symmetric Equations

When you have \(x+y=S\) and \(x^2+y^2=V\), you are essentially solving a quadratic equation. The roots of \(t^2 - (x+y)t + xy = 0\) will give you \(x\) and \(y\). Here, we found \(xy\) using the identity \((x+y)^2\), leading to the quadratic \(t^2 - 14t + 48 = 0\), whose roots are clearly 8 and 6.

3. Probability without Replacement

When picking "one after another without replacement", the order matters (Permutations). However, for problems involving the "minimum" or "maximum" of a set, you can often use Combinations (\(^nC_r\)) because the set of two numbers \(\{2, 3\}\) has the same minimum regardless of whether 2 was picked first or 3 was picked first.

4. The Power of Complementary Events

If a question asks for "at least one", "less than", or "smaller is less than", always check if the opposite case is easier to count. Here, "smaller is less than 4" covers many cases (1, 2, or 3 being picked). The opposite, "both are 4 or greater", is a much smaller subset to calculate.

5. JEE Exam Relevance

Statistics usually provides 1-2 questions per shift in JEE Main. These are considered "easy-scoring" marks. Probability is more varied, but combining it with Statistics is a common tactic to increase the difficulty level from a simple formula-based question to a multi-step logical problem.

6. Common Mistakes

  • Calculation Errors: Incorrectly squaring 12 or 14 is the most common pitfall in variance problems.
  • Replacement Confusion: Treating the problem as "with replacement" would change the denominator to \(6 \times 6 = 36\).
  • Order in Set: Misidentifying \(x-4\) or \(y\) in the second set before solving Step 3.

7. Shortcut: Mean Deviation vs Variance

While Mean Deviation is \(\frac{\sum |x_i - \bar{x}|}{n}\), Variance uses squares to eliminate the sign. Squares penalize outliers more heavily, which is why the Variance in this problem (16) is quite high relative to the mean (8).

8. Properties of Variance

  • \(Var(ax + b) = a^2 Var(x)\).
  • Variance is always non-negative.
  • If all observations are increased by a constant \(k\), variance remains unchanged.

9. Probability Sample Space

For a set of size \(n\), choosing 2 elements without replacement results in \(n(n-1)\) ordered pairs or \(\frac{n(n-1)}{2}\) unordered pairs. Consistency is key—if you use combinations for the numerator, use them for the denominator as well.

10. Summary for Students

To master such problems: 1. Memorize the algebraic relationship between Mean, Variance, and \(\sum x^2\). 2. Practice solving quadratics from sum and product. 3. Always look for the complement in probability to save time.

Related JEE Main Questions

Frequently Asked Questions (FAQs)

1. What is the variance formula used here? We used \(\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2\), which is standard for finding missing data points.
2. Why did we assume x and y are 8 and 6? Because their sum is 14 and product is 48. Solving \(t^2 - 14t + 48 = 0\) gives 8 and 6. Since \(x > y\), \(x=8\).
3. What does "without replacement" mean? It means once a number is picked, it cannot be picked again for the second draw.
4. How many elements are in the second set? There are 6 elements: {1, 2, 3, 4, 6, 5}.
5. Why use the complement method for probability? Calculating "smaller is less than 4" involves many cases. "Both are 4 or more" only has 3 possible numbers {4, 5, 6}, making it much faster.
6. Is order important in the probability calculation? Since it says "one after another", order is technically important, but the final probability ratio remains the same whether you use permutations or combinations.
7. What happens to variance if all numbers are doubled? The variance would become \(2^2 = 4\) times the original variance.
8. What is the sum of squares of the first n natural numbers? Though not used here, it is \(\frac{n(n+1)(2n+1)}{6}\), often used in other JEE statistics problems.
9. Can x and y be fractions? In this specific quadratic equation, the roots are integers, which is common for JEE Main level.
10. Why is the total sample space 30? Picking 2 from 6 without replacement is \(6 \times 5 = 30\).
J

Expert Contribution by: JEE NEET Experts

Jee neet experts, 10 year experience | Specializing in Statistics and Probability.

Related Covered Topics

jee mainsstatisticsvariance formulamean and varianceprobability without replacementiit jee mathssample spacecomplementary eventsalgebraic identitiesquadratic equationsdescriptive statisticscombinatoricspermutation and combinationjee 2026math shortcutsengineering entrance exammissing observationsdata analysisiit entrance tipsadvanced math problems
Scroll to Top