Probability problem with tissues Girl A has 10 new tissues and 5 used tissues in her drawer. Girl B comes to visit girl A. Girl A makes her friend a coffee and randomly takes out 2 tissues out of the drawer. Later she puts them back into the drawer. The next day she takes out 2 tissues again. What is the probability that both of the tissues are used? 
The answer is 359/2205 . I've been trying to solve this without any luck.

Initially we have ten unused tissues and five used ones. There are three different use cases:


*

*The girl takes out two used tissues on the first day. Both the first and the second drawn tissue must be used, so this probability equals $\frac{5}{15} \cdot \frac{4}{14}$;

*The girl takes out one used and one unused tissue on the first day, which comes down to either taking first one used and then one unused tissue, or taking first one unused and then one used tissue. As such, the probability of this happening equals $\frac{10}{15} \cdot \frac{5}{14} + \frac{5}{15} \cdot \frac{10}{14}$;

*The girl takes out two unused tissues on the first day, which has a probability of $\frac{10}{15} \cdot \frac{9}{14}$.


The probability of taking out two used tissues on the second day equals $\frac{5}{15} \cdot \frac{4}{14}$ in the first case, $\frac{6}{15} \cdot \frac{5}{14}$ in the second and $\frac{7}{15} \cdot \frac{6}{14}$ in the third. Overall, we get:
$$P[2\,used\,tissues] = \Big(\frac{5}{15} \cdot \frac{4}{14}\Big) \Big(\frac{5}{15} \cdot \frac{4}{14}\Big) + \Big(\frac{10}{15} \cdot \frac{5}{14} + \frac{5}{15} \cdot \frac{10}{14}\Big) \Big(\frac{6}{15} \cdot \frac{5}{14}\Big) + \Big(\frac{10}{15} \cdot \frac{9}{14}\Big) \Big(\frac{7}{15} \cdot \frac{6}{14}\Big) = \frac{7180}{44100} = \frac{359}{2205}$$
