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JEE Main 202621 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Statistics

JEE Main 2026 · 21 January · Morning Shift · Mathematics · Statistics

Let the mean and variance of 7 observations 2,4,10, x, 12,14, y,...

Let the mean and variance of 7 observations 2,4,10,x,12,14,y,x>y2,4,10, x, 12,14, y, x>y, be 8 and 16 respectively. Two numbers are chosen from {1,2,3,x−4,y,5}\{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 :

Options

  1. A35\frac{3}{5}
  2. B13\frac{1}{3}
  3. C25\frac{2}{5}
  4. D45\frac{4}{5}

Correct answer

D. 45\frac{4}{5}

Step-by-step solution

Mean = 8 gives x+y=14x + y = 14.
Variance = 16 gives (x−8)2+(y−8)2=4(x-8)^2 + (y-8)^2 = 4.
Solving: x=8x = 8, y=6y = 6 (since x>yx > y).
Set becomes {1,2,3,4,6,5}={1,2,3,4,5,6}\{1, 2, 3, 4, 6, 5\} = \{1, 2, 3, 4, 5, 6\}.
Total ways to choose 2 numbers = (62)=15\binom{6}{2} = 15.
Favorable cases (smaller <4< 4): smaller = 1 gives 5 pairs, smaller = 2 gives 4 pairs, smaller = 3 gives 3 pairs.
Total favorable = 12.
Probability =1215=45= \frac{12}{15} = \frac{4}{5}.
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