JEE Main 2026 — 21 January, Morning Shift — Question 13

Mean $(\overline{\mathrm{x}})=8$ (Given)

$\Rightarrow \frac{2+4+10+x+12+14+y}{7}=8$

$\Rightarrow \mathrm{x}+\mathrm{y}=14 \ldots\ldots$ Variance $\left(\sigma^{2}\right)=16$ (Given)

$\Rightarrow 16=\frac{2^{2}+4^{2}+10^{2}+\mathrm{x}^{2}+12^{2}+14^{2}+\mathrm{y}^{2}}{7}-8^{2}$

$\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}=100$ (2).....

$\because(x+y)^{2}=x^{2}+y^{2}+2 x y$

$\Rightarrow \mathrm{xy}=48$ (sum is 14 product is 48 )

Since problem states $x>y$

$\therefore \mathrm{x}=8$ and $\mathrm{y}=6$

Now set $X=\{1,2,3,4,6,5\}$

Now we choose two numbers one after author without replacement total outcomes $=6 \times 5=30$

We want the prob.

That the smaller number among the two is less than 4 $\mathrm{P}($ smaller $<4)=1-\mathrm{P}($ smaller $\geq 4)$ $=1-\frac{6}{30}=\frac{4}{5}$