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# In a non-leap year,the probability of having 53 tuesdays or 53 wednesdays is

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

$\therefore$ A non leap year has 365 days
$\therefore 52\times 7=364$
$\therefore$ Remaining is 1 day which could be either of the day out of 7 days.
$\therefore$ Probability of having 53 tuesdays or 53 wednesdays =$\large\frac{1}{7}$