Solution

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For median to be greater than 2 there can be two cases

• Case 1. all three numbers are greater than 2.
  
  

  \( p(case \ 1) = (\frac{3-2}{3-0})^3 = \frac{1}{27} \)

• Case 2. two numbers are >=2 while one number is <= 2
  
  

  \( p (case \ 2) = \binom{3}{2} * (\frac{1}{3})^2 * \frac{2}{3} = \frac{2}{9} \)

final answer will be sum of these two probabilities \(=\frac{7}{27} \)