Browse Questions

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

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

Can you answer this question?

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.
answered Jul 8, 2014