$\begin{array}{1 1}(A)\;\large\frac{1}{7}\\(B)\;\large\frac{2}{7}\\(C)\;\large\frac{3}{7}\\(D)\;\text{None of these}\end{array} $

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Step 1:

Given a non-leap year

$\therefore$ A non leap year has 365 days

Which includes 52 whole weeks

$\therefore 52\times 7=364$

Step 2:

$\therefore$ Remaining is 1 day which could be either of the day out of 7 days.

$\therefore$ Probability of having 53 tuesdays or 53 wednesdays =$\large\frac{1}{7}$

Hence (A) is the correct answer.

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