<s>[INST] Probability of exactly $2$ sixes in $3$ dice rolls where $2$ dice have $6$ on $2$ faces? 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 ?
 [/INST] Your solution seems correct, indeed also by the naif definition of probability we obtain
$$p=\frac{\text{#favorauble cases}}{\text{#total cases}}=\frac{8+8+20}{6^3}=\frac16$$
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