Gate2020 cs Q49

  • Option : A
  • Explanation :
    Let n = 3
    0 ϵ < 0,1]3
    Gate2020 cs
    x = 0 or 1
    Gate2020 cs
    = x1 = x3 (∵ a1 = 1, a2= 0, a3 = 1)
    = 0 or 1 or 1 or 2.
    Gate2020 cs
    Gate2020 cs
    = x3 (∵ a1 = a2 = 0)
    = 0 or 1
    Gate2020 cs
    Total number of cases = 2 × 3 + 4 × 3 = 18
    Now odd number of cases = 9
    P = 9/18 = 0.5
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