Let $A = \{\text{sum of digits is }8\}, \qquad B = \{\text{product of digits is }0\}.$ We need the conditional probability $P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{n(A\cap B)}{n(B)}.$

The tickets are numbered from $00$ to $49$, so there are $50$ equally likely tickets.

First count $B$: product of digits is $0$.

A two-digit ticket has tens digit from $0$ to $4$ and units digit from $0$ to $9$.

The product is $0$ when at least one digit is $0$.

• Tens digit $0$: tickets $00,01,02,\dots,09$ gives $10$ tickets.
• Units digit $0$: tickets $00,10,20,30,40$ gives $5$ tickets.
• But $00$ is counted twice, so subtract $1$.

Hence, $n(B)=10+5-1=14.$

Now count $A\cap B$: sum of digits is $8$ and product is $0$.

Since product is $0$, one digit must be $0$. Since sum is $8$, the other digit must be $8$.

So the possible tickets are $08$ and $80$.

But the tickets only go up to $49$, so $80$ is not available. Therefore only $08$ works.

Hence, $n(A\cap B)=1.$

Therefore, $P(A\mid B)=\frac{1}{14}.$

Comparing with the options, the correct option is $\boxed{\text{D } \frac{1}{14}}.$