Question

What is the probability that a non-leap year has 53 Fridays?

• A. $2/7$
• B. $3/7$
• C. $1/7$
• D. $5/7$

Answer 1

The correct option is C $1/7$

There are $365$ days in a leap year $=52$ weeks $+$ $1$ day.
$52$ weeks have $52$ Fridays.
So, the last $1$ day can be any of the seven days of the week out of which only one day is Friday.
So, Probability that a non leap yeas has $53$ Fridays$=\frac{Numberoffavourableoutcomes}{Totalnumberofoutcomes}=\frac{1}{7}$.