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.
 A: The first time girl B chooses two tissues, there are three possibilities:
a) Both of the tissues are used:   Probability: $\dfrac{\binom{5}{2}}{\binom{15}{2}}=\dfrac{2}{21}$
b) Both of the tissues are not used: Pobability: $\dfrac{\binom{10}{2}}{\binom{15}{2}}=\dfrac{9}{21}$
c) One of the tissues is used and one of them is not: Probability:$\dfrac{\binom{5}{1}\binom{10}{1}}{\binom{15}{2}}=\dfrac{10}{21}$
If situation "a" happens, the second time we want to choose two tissues, there are still 5 used tissues. So the probability that both of them are used is $\dfrac{\binom{5}{2}}{\binom{15}{2}}=\dfrac{2}{21}$. So if we want to calculate the probability of situation, we have to multiply this number, by the probability of a: $\dfrac{2}{21}\times\dfrac{2}{21}=\dfrac{4}{441}$ 
If situation "b" happens, we have 7 used tissues and if situation "c" happens we will have 6 used tissues. Now the same way, you can yield these numbers for situation "b" and "c":$\dfrac{9}{105}$ and $\dfrac{10}{147}$
Now we have to add these fractions: $\dfrac{4}{441}+\dfrac{9}{105}+\dfrac{10}{147}=\dfrac{359}{2205}$
A: I assume that on the first day, both of the tissues taken out become used (if they weren't already).
For the first day, the probability that she takes out two new tissues is $\frac{10}{15}\times\frac{9}{14}$. If this happens, she now has 8 new and 7 used on the second day. So you can work out the probability of taking out two used tissues on the second day given that she took out two new tissues on the first day. Then multiply these two probabilities together to get the probability that she took out two new tissues on the first day and then two used tissues on the second.
Once you've done that you need to work out, in a similar way, the probability she takes out one of each on the first day and then two used on the second, and the probability that she takes out two used on both days. 
A: 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:
$$P[2\,used\,tissues] = \Big(\frac{5}{15} \cdot \frac{4}{14}\Big) \Big(\frac{5}{15} \cdot \frac{4}{14}\Big) + \Big(\frac{10}{15} \cdot \frac{5}{14} + \frac{5}{15} \cdot \frac{10}{14}\Big) \Big(\frac{6}{15} \cdot \frac{5}{14}\Big) + \Big(\frac{10}{15} \cdot \frac{9}{14}\Big) \Big(\frac{7}{15} \cdot \frac{6}{14}\Big) = \frac{7180}{44100} = \frac{359}{2205}$$
