Probability of exactly $2$ sixes in $3$ dice rolls where $2$ dice have $6$ on $2$ faces?

Question

Three dice are rolled. One is fair and the other two have 6 on two faces.

Find the probability of rolling exactly 2 sixes.

My textbook gives an answer of $\frac{20}{147}$ but I get an answer of: $$\frac{1}{6}\frac{2}{6}\frac{4}{6}+\frac{1}{6}\frac{4}{6}\frac{2}{6}+\frac{5}{6}\frac{2}{6}\frac{2}{6}=\frac{8}{216}+\frac{8}{216}+\frac{20}{216}=\frac{36}{216}=\frac{1}{6}$$

I just want to know where I am going wrong or could the textbook be mistaken ?

Answer 1

Note that the official answer is correct if you make the (unnatural) assumption that the two non-standard dice have seven sides (two of which show $6$).

In that case the answer is $$\frac 27\times \frac 27\times \frac 56+2\times \frac 27\times \frac 57\times \frac 16=\frac {20}{147}$$

To be sure, this was arrived at by reverse engineering, not by any sensible reading of the problem.

Answer 2

Your solution seems correct, indeed also by the naif definition of probability we obtain

$$p=\frac{\text{#favorauble cases}}{\text{#total cases}}=\frac{8+8+20}{6^3}=\frac16$$