Gate2019 cs Q56

  • Option : A
  • Explanation :
    No. of pairs with path length 0=8.0=8.
    No. of pairs with path length 1=0.1=0.
    No. of pairs with path length 2=8.2=8.
    No. of pairs with path length 3=0.3=0.
    No. of pairs with path length 4=16.4=16.
    No. of pairs with path length 5=0.5=0.
    No. of pairs with path length 6=32.6=32.
    Total number of possible pairs =8×8=64=8×8=64
    So, expected path length,E(x),
    =0×864+2×864+4×1664+6×3264=27264=4.25
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