Five captains are arranged randomly in a line. The total number of possible orderings is \[ 5! = 120. \] Treat the India and Australia captains as a single block to ensure they stand next to each other. Then we have this block plus the other three captains, i.e. \(4\) items, which can be arranged in \(4!\) ways. Inside the block the two captains can be ordered in \(2!\) ways. Hence the number of favourable arrangements is \[ 2!\cdot 4! = 2\cdot 24 = 48. \] Therefore the required probability is \[ \frac{48}{120}=\frac{2}{5}. \] \[ \boxed{\dfrac{2}{5}} \]