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

Let A be the event that the sum of the digits on the selected ticket is 8.
Let B be the event that the product of the digits on the selected ticket is 0.
The tickets numbered from 00 to 49 are selected.
The numbers whose sum of digits is 8 are 08, 17, 26, 35, 44. There are 5 such numbers.
A = {08, 17, 26, 35, 44}
The numbers whose product of digits is 0 are those with at least one digit being 0. These numbers are 00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40.
B = {00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40}
There are 14 such numbers.
The numbers whose sum of digits is 8 and product of digits is 0 are 08.
Thus, A ∩ B = {08}
There is only 1 such number.
P(A|B) = P(A ∩ B) / P(B) = (1/50) / (14/50) = 1/14