One ticket is selected at random from 50 tickets numbered 00, 01, 02, ..., 49. Then the probability that the sum of the digits on the selected ticket is 8, given that the product of these digits is zero, equals

$\frac{1}{14}$

Consider the following events:
A = Sum of the digits on the selected tickets is 8.

B= Product of the digits on the selected ticket is zero.

There are 14 tickets having product of digits appearing on them as zero. The numbers on such tickets are 00, 01, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40.

$∴P(B)=\frac{14}{50}$ and $P(A ∩ B) =\frac{1}{50}$

Required probability $=P(A/B)=\frac{P(A ∩ B)}{P(B)}=\frac{1}{14}$