Question

The probability that a non-leap year has 53 Sundays is ______.

Options

• 1

• \[\frac{1}{7}\]

• \[\frac{1}{365}\]

• \[\frac{7}{365}\]

Answer 1

The probability that a non-leap year has 53 Sundays is \[\underline{\boldsymbol{\frac{1}{7}}}\].

Explanation:

There are 365 days in a non-leap year.

No. of weeks = \[\frac{365}{7} = 52\] weeks and 1 day.

Since, there is one sunday in a week, and one week has seven days.

So, in order to have 53 sundays the 1 day that remains should also be sunday.

P(that a non-leap year has 53 sundays)

\[{} = \frac{\text{No. of favourable outcomes}}{\text{Total no. of possible outcomes}} = \frac{1}{7}\]