Question

If a leap-leap year selected at random, the chance that it will contain $53$ Sundays is

• A. $\frac{3}{7}$
• B. $\frac{1}{7}$
• C. $\frac{1}{367}$
• D. None of these

Answer 1

The correct option is A $\frac{3}{7}$

A leap-leap year has $367$ days i.e. $52$ complete week and three days more. These three days will be three consecutive days of a week. A leap-leap year will have $53$ Sundays if out of the three consecutive days of a week selected at random one is a Sunday.
Let $S$ be the sample space and $E$ be the event that out of the three consecutive days of a week one is Sunday, then
$S=$ {(Sunday,Monday,Tuesday,(Monday,Tuesday,Wednesday),(Tuesday,Wednesday,Thursday),(Wednesday,Thursday,Friday),(Thursday,Friday,Saturday),(Friday,Saturday,Sunday),(Saturday,Sunday,Monday)}
$\therefore n\left(S\right)=7$
and $E=${(Sunday,Monday,Tuesday),((Friday,Saturday,Sunday),(Saturday,Sunday,Monday)}
$\therefore n\left(E\right)=3$
Hence, Required probability $P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{3}{7}$