Son whose name starts with 'M'

We are asked to find $P(\textrm{2 sons }$ $|$ $\textrm{ son with letter M})$ . Using Bayes theorem:

$P(\textrm{2 sons }$ $|$ $\textrm{ son with letter M})=\frac{P(\textrm{son with letter M}|\textrm{2 sons})P(\textrm{2 sons})}{P(\textrm{son with letter M})}$

We have $P(\textrm{son with letter M}$ $|$ $\textrm{2 sons})=\frac{2}{26}-(\frac{1}{26})^{2}=\frac{51}{26^{2}}$

$P(\textrm{2 sons})=\frac{1}{2}^{2}=\frac{1}{4}$

and $P(\textrm{son with letter M})=\frac{2}{52}-(\frac{1}{52})^{2}=\frac{103}{52^{2}}$

Plugging all this in, we obtain $P(\textrm{2 sons }$ $|$ $\textrm{ son with letter M})=\frac{(\frac{51}{26^{2}})(\frac{1}{4})}{\frac{103}{52^{2}}}=\frac{51}{103}$

Therefore the answer is $51+103=\boxed{154}$

Consider the number of ways Alice can have two children whose names start with given letters.

If Alice has an older son and a younger daughter, the son's name must start with 'M' while the daughter's name might start with anything. Thus, there are 26 ways that Alice could have an older son and a younger daughter.

If Alice has an younger son and an older daughter, the son's name must start with 'M' while the daughter's name might start with anything. Thus, there are 26 ways that Alice could have a younger son and an older daughter.

If Alice has two sons, then there are 25 ways that the older son's name starts with 'M' and the younger's does not, 25 ways that the younger son's name starts with 'M' and the older's does not, and just 1 way for them both to have names that start with 'M'. Thus, there are 51 ways that Alice could have two sons.

Finally, Alice cannot have two daughters.

Therefore, the probability that Alice has two sons is $\frac{51}{26+26+51}=\frac{51}{103}$ , so the answer is $\boxed{154}$ .