We can solve this problem using Bayes' Theorem. $$ P(A|B) = \frac{P(B|A) P(A))}{P(B)} $$ Here, $A$ is the event that the coin I chose was the double-header, and $B$ is the event that I tossed the chosen coin three times and it landed $H$ every time. The probability of choosing the double header is $$ P(A) = 0.1 $$ and the probability of tossing three heads in a row is $$ P(B) = 0.9 \left( \frac{1}{2^3} \right) + 0.1 (1) = \frac{17}{80} $$ since with $P = 0.9$ we choose a regular coin which will throw three heads with probability $\frac{1}{2^3}$, and with $P = 0.1$ we choose a coin which will certainly land $H$ every time. What remains is to calculate $P(B|A)$, the probability of throwing three heads given we have the double-headed coin. We will always throw three heads with the double-header and therefore $$ P(B|A) = 1 $$ Now, we can calculate $P(A|B)$ $$ P(A|B) = \frac{0.1}{\left( \frac{17}{80} \right)} = \frac{8}{17} \approx 0.47. $$