Statistics – Mean and Variance – JEE Main 21 Jan 2026 Shift 1

Solution:

Given 7 observations: $2, 4, 10, x, 12, 14, y$.
Mean $\bar{x} = 8$.
$$\frac{2+4+10+x+12+14+y}{7} = 8$$
$$42 + x + y = 56 \Rightarrow x + y = 14$$

Variance $\sigma^2 = 16$.
$$\frac{\sum x_i^2}{n} – (\bar{x})^2 = 16$$
$$\frac{2^2 + 4^2 + 10^2 + x^2 + 12^2 + 14^2 + y^2}{7} – 8^2 = 16$$
$$\frac{4 + 16 + 100 + x^2 + 144 + 196 + y^2}{7} = 16 + 64 = 80$$
$$460 + x^2 + y^2 = 560$$
$$x^2 + y^2 = 100$$

We have a system of equations:
1) $x + y = 14$
2) $x^2 + y^2 = 100$
Using $(x+y)^2 = x^2 + y^2 + 2xy$:
$$196 = 100 + 2xy \Rightarrow 2xy = 96 \Rightarrow xy = 48$$
Since given $x > y$, we get $x = 8$ and $y = 6$.

The set of numbers to choose from is $S = \{1, 2, 3, x-4, y, 5\}$.
Substitute $x=8, y=6$:
$$S = \{1, 2, 3, 4, 6, 5\} = \{1, 2, 3, 4, 5, 6\}$$
Total elements = 6.

Experiment: Two numbers are chosen without replacement.
Total outcomes = $6 \times 5 = 30$.

We need $P(\text{smaller number} < 4)$. It is easier to find $P(\text{smaller number} \ge 4)$ and subtract from 1. For the smaller number to be $\ge 4$, both numbers must be chosen from the subset $\{4, 5, 6\}$. Number of favorable outcomes for complement = $3 \times 2 = 6$. $$P(\text{smaller} \ge 4) = \frac{6}{30} = \frac{1}{5}$$
Required Probability = $1 – P(\text{smaller} \ge 4)$
$$= 1 – \frac{1}{5} = \frac{4}{5}$$

Ans. (2)