Two fair dice are rolled simultaneously. One of the dice shows four. The probability of other dice showing six, is equal to

$\frac{2}{11}$

Following equally likely outcomes may occur when one of the dice show four;

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

Out of these eleven outcomes exactly 2 outcomes favor the cause of second dice showing a six.

Thus, required probability = $\frac{2}{11}$