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# The probability that a non - leap year has 53 sundays is

$\begin {array} {1 1} (A)\;\large\frac{1}{7} & \quad (B)\;\large\frac{2}{7} \\ (C)\;\large\frac{3}{7} & \quad (D)\;\large\frac{4}{7} \end {array}$

Can you answer this question?

A non leap year contains 365 days.
i.e., 52 weeks and one day
This one day can be any one of seven days of the week.
Required probability = $\large\frac{1}{7}$
answered Jan 6, 2014