Question

What is the chance that a non-leap year contains $53$ saturdays?

• A. $\frac{2}{7}$
• B. $\frac{1}{7}$
• C. $\frac{2}{365}$
• D. $\frac{1}{365}$

Answer 1

The correct option is B $\frac{1}{7}$

A non-leap year has $365days$

A year has $52weeks.$ Hence there will be 52 Sundays for sure.

$52weeks=52x7=364days.$

$365–364=1dayextra.$

In a non-leap year there will be 52 Sundays and 1day will be left.

This 1 day can be Sunday, Monday, Tuesday, Wednesday, Thursday,friday,Saturday, Sunday.

Of these total 7 outcomes, the favourable outcomes are 1.

Hence the probability of getting 53 sundays $P\left(E\right)=\frac{1}{7}$