Geometric conditional Probability In order to start a game, each player takes turns throwing a fair six-sided dice until a $6$ is obtained. Let $X$ be the number of turns a player takes to start the game. Given that $X=3$, find the probability that the total score on all three of the dice is less than $10$.
So initially I thought this was straight forward. Total number of outcomes that three dice can sum up to less than $10$ is equal to $60$. Total amount of outcomes is $6^3=216$.
Therefore probability of the total summing to less than $10$ is $$P(\text{sum} < 10)=\frac{60}{6^3}=\frac{5}{18}$$.
However, this did not match the provided solution of $1/12$ and so I thought it had something to do with conditional probability. i.e. $P(\text{sum} < 10 | X = 3)$.
So naturally, $$P(\text{sum} < 10 | X = 3)=\frac{P(X = 3 | \text{sum} < 10) P(\text{sum} < 10)}{P(X=3)}$$
Out of the $60$ outcomes that sum less than $10$, only $6$ of those outcomes contain sixes. So that when we have to calculate the probability of taking three turns to get one $6$ we have to fail twice and succeed once and hence$$P(X = 3 | \text{sum} < 10) = \left(\frac{9}{10}\right)^2\left(\frac{1}{10}\right)=\frac{81}{1000}$$
$P(X=3)$ is simply $\left(\frac{5}{6}\right)^2\left(\frac{1}{6}\right)=\frac{25}{216}$ and so finally
$$P(\text{sum} < 10 | X = 3)=\frac{P(X = 3 | \text{sum} < 10) P(\text{sum} < 10)}{P(X=3)}=\frac{\frac{81}{1000} \frac{5}{18}}{\frac{25}{216}}=\frac{243}{1250}$$
Where did I go wrong? Alternatively, maybe the textbook is incorrect?

For a more brute force approach, consider the following. You have three numbers, you know the last one is 6. There are $6 \cdot 6=36$ possible combinations for the first two numbers (assuming they can be any number from 1 to 6, which your problem seems to state that they cannot). Asking that the sum of these three numbers is $<10$ is equivalent to asking that the sum of the first two numbers is $<4$. Counting it all out you find there are 3 such combinations. Therefore $3/36=1/12$.
