Statistics · Question 16

JEE Main 2026 (Online) 21st January Morning ShiftMCQMedium📈 Topic in 45% papers+4−1

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

(1) 45

Solution

Mean = 8 2 + 4 + 10 + x + 12 + 14 + y = 56

x + y = 14 ...(i)

Variance = 16 xi27 - 82 = 16

xi2 = 560

4 + 16 + 100 + 144 + 196 + x2 + y2 = 560

x2 + y2 = 100 ...(ii)

From (i) and (ii): 2xy = 142 - 100 = 96 xy = 48

x = 8, y = 6 (since x > y)

Set = \1, 2, 3, 4, 6, 5\, i.e. the numbers 1 to 6.

Smaller number 4 only when both chosen numbers are from \4, 5, 6\.

P(smaller 4) = 3C26C2 = 315 = 15

P(smaller < 4) = 1 - 15 = 45

Answer: (1)

JEE Main 2026 (Online) 21st January Morning Shift

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