JEE Main 2026 — Jan 21, Shift 1 — Mathematics solution

JEE Main2026Jan 21, Shift 1Mathematics

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
$\frac{3}{5}$
B
$\frac{4}{5}$
C
$\frac{2}{5}$
D
$\frac{1}{3}$

Solution

Mean $= 8$ : $(2 + 4 + 10 + 12 + 14) + x + y = 56 \Rightarrow x + y = 14$ .

Variance $= 16$ : $\frac{\sum x _{i}^{2}}{7} - 64 = 16 \Rightarrow \sum x_{i}^{2} = 560$ . Removing the fixed values: $x^{2} + y^{2} = 560 - 460 = 100$ .

From $(x + y)^{2} = x^{2} + y^{2} + 2 x y$ : $196 = 100 + 2 x y \Rightarrow x y = 48$ . Since $x > y$ : $x = 8$ , $y = 6$ .

The set becomes ${1, 2, 3, 4, 6, 5}$ (since $x - 4 = 4$ ). Total ordered draws $= 6 \times 5 = 30$ .

The smaller is $\geq 4$ iff both picks are from ${4, 5, 6}$ : $3 \times 2 = 6$ ways. So $P (smaller < 4) = 1 - 6/30 = 4/5$ .

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