The probability that a leap year contains 53 Modanys and 53 Tuesdays is

The number of days in a leap year is 366

Leap year have 52 weeks and 2 more days

To have 53 Mondays and 53 Tuesdays in a leap year, the extra two days must be Monday and Tuesday

The possibiliites are $$\begin{gathered} S=\left\{\left(Sun,Mon\right),\left(Mon,Tue\right),\left(Tue, Wed\right), \\ \left(Wed,Thu\right),\left(Thu,Fri\right),\left(Fri,sat\right),\left(Sat,Sun\right)\right\} \end{gathered}$$

hence, $n\left(S\right)=7$

The number of favorable cases to the evennt of getting two days must be Monday and Tuesday is $1:\left\{\left(Mon,Tue\right)\right\}$

Therefore,the probability that a leap year contains 53 Modanys and 53 Tuesdays is $\boxed{\frac{1}{7}}$