smc

The digit $7$ can be either the tens digit ($70,71,\dots ,79$: $10$ possibilities), or the ones digit ($17,27,\dots ,97$: $9$ possibilities), but we counted the number $77$ twice. This means that out of the $90$ two-digit numbers, $10+9-1=18$ have at least one digit equal to $7$. Therefore the probability is $\frac{18}{90}=\boxed{\frac{1}{5}}⟹(\mathrm{B})$.