Federer, Nadal, Djokovic and Murray are the four players left in a singles tennis tournament. They are randomly assigned opponents in the semi final matches, and the winners of those matches play each other in the final match to determine the winner of the tournament. When Federer plays Nadal, Federer will win the match with probability $\frac{2}{3}$. When either Federer or Nadal plays either Djokovic or Murray, Federer or Nadal will win the match with probability $\frac{3}{4}$. Assume that outcomes of different matches are independent. The probability that Nadal will win the tournament is $\frac{p}{q}$, where p and q are relatively prime positive integers. Find units digit of $p+q.$

Let F be Federer, N be Nadal, J be Djokovic, and M be Murray. The 4 circles represent the 4 players, and the arrow is from the winner to the loser with the winning probability as the label.
 
The problem can be solved in 2 cases.
Case1: N’s opponent for the semifinals is F.
The probability N’s opponent is F is $\frac{1}{3}$ . Therefore the probability N wins the semifinal in this case is  $\frac{1}{3}.\frac{1}{3}$. The other semifinal game is played between J and M, it doesn’t matter who wins because N has the same probability of winning either one. The probability of N winning in the final is $\frac{3}{4}$ , So the probability of N winning the tournament in case1 is  $\frac{1}{3}.\frac{1}{3}.\frac{3}{4}$
Case2: N’s opponent for the semifinal is J or M.
It doesn’t matter if N’s opponent is J or M because N has the same probability of winning either one. The probability N’s opponent is J or M is $\frac{2}{3}$ . Therefore the probability N wins the semifinal in this case is $\frac{2}{3}.\frac{3}{4}$. The other semifinal game is played between F and J or M. In this case it matters who wins in the other semifinal game because the probability of N winning F and J or M is different.
Case2.1: N’s opponent for the final is F.
For this to happen, F must have won J or M in the semifinal, the probability is  $\frac{3}{4}$. Therefore the probability that N won F in the final is  $\frac{3}{4}.\frac{1}{3}$
Case2.2: N’s opponent for the final is J or M.
For this to happen, J or M must have won F in the semifinal, the probability is $\frac{1}{4}$. Therefore the probability that N won J or M in the final is  $\frac{1}{4}.\frac{3}{4}$
In Case2, the probability of N winning the tournament is  $\frac{2}{3}.\frac{3}{4}.(\frac{3}{4}.\frac{1}{3}+\frac{1}{4}.\frac{3}{4})$
Adding Case1 and Case2 together, we get $\frac{1}{3}.\frac{1}{3}.\frac{3}{4}+\frac{2}{3}.\frac{3}{4}.(\frac{3}{4}.\frac{1}{3}+\frac{1}{4}.\frac{3}{4})=\frac{29}{96}$,
So the answer is 29 + 96 = 125.