Question

Find the probability of getting $53$ sundays in a non leap year.

• A. $\frac{2}{7}$
• B. $\frac{1}{7}$
• C. $\frac{1}{5}$
• D. $\frac{1}{9}$

Answer 1

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

A non-leap year has 365 days

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

52 weeks = 52 x 7 = 364 days .

365– 364 = 1 day extra.

In a non-leap year there will be 52 Sundays and 1 day 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 = 1 / 7.