(A) \(\frac{4}{5}\)
(B) \(\frac{3}{5}\)
(C) \(\frac{2}{5}\)
(D) \(\frac{1}{3}\)
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.
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:
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.
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.
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.
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.
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.
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).
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.
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.
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