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 $\{1,2,3, \mathrm{x}-4, \mathrm{y}, 5\}$ one after another without replacement, then the probability, that the smaller number among the two chosen numbers is less than 4 , is:

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Statistics

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:

Answer: (B)

The mean gives

x + y = 14

and the variance gives

x^{2} + y^{2} = 100

, hence

x y = 48

; with

x > y

this forces

x = 8, y = 6

. The set

{1, 2, 3, x - 4, y, 5}

becomes

{1, 2, 3, 4, 5, 6}

. Choosing two numbers one after another gives

6 \times 5 = 30

ordered outcomes. The smaller of the two is at least

4

only when both come from

{4, 5, 6}

, i.e.

3 \times 2 = 6

outcomes. Therefore the probability that the smaller number is less than

4

1 - \frac{6}{30} = \frac{4}{5}

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Single correct option