Two fair dice are rolled simultaneously. It is found that one of them shows odd prime numbers. The probability that remaining dice also shows an odd prime number, is equal to

$\frac{1}{5}$

3 and 5 are the only odd prime numbers, among the possible outcomes. The following are the outcomes when one of them show odd prime numbers,

(3, 1), (1, 3) (3, 2), (2, 3) (3, 3), (3, 4) (4, 3), (3, 5) (5, 3), (3, 6) (6, 3), (5, 1), (1, 5), (5, 2), (2, 5), (5, 4), (4, 5), (5, 5), (5, 6), (6, 5).

Out of these 20 equally likely outcomes exactly 4 favour the presence of odd prime numbers on both dice.

Thus, required probability = $\frac{4}{20} = \frac{1}{5}$