Question

In a leap year the probability of having 53 Sundays and 53 Monday is
(a) $\frac{1}{7}$
(b) $\frac{3}{7}$
(c) $\frac{4}{7}$
(d) $\frac{5}{7}$

Answer 1

The correct option is (A): $\frac{1}{7}$

A leap year has 366 days i.e ($52\times 7+2$ days)= 52 weeks + 2 days

Since 364 is divisible by 7

Hence, two excess weekdays can be

i.e. S = {(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)}

i.e. 7 pairs of excess weekdays.

Let E be the event of having 53 Sundays and 53 Mondays.

i.e E ={(Sunday, Monday)}

∴ The probability that a leap year will contain 53 Sundays and 53 Mondays.

$=\frac{n\left(E\right)}{n\left(S\right)}=\frac{1}{7}$