The probability of a leap year having 5 Sundays in February is-
A. 0
B. 1
C.$\dfrac{1}{7}$
D.$\dfrac{2}{7}$

Hint: A leap year has 366 days in a year and 29 days in February month. Which means that in February month there are 4 weeks and 1 day extra. So, this extra day can be any day of the 7 days of the week. It could be any of these 7 days-
Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday.


Complete step by step solution
Given:
Let us suppose that the name of this event is $A$.
We know that the total number of possible outcomes of this event is $n = 7$ and number of favorable outcomes of this event (i.e. the day being Sunday) is,
$n\left( A \right) = 1$
Since, we know that from the formula of probability that the Probability of a day being Sunday out of these 7 days is,
 $P\left( A \right) = \dfrac{{n\left( A \right)}}{n}$
On substituting the value of $n\left( A \right)$ as 1 and $n$ as 7 in the above expression, we get,
$P\left( A \right) = \dfrac{1}{7}$
So, the probability of a leap year having 5 Sundays in February is $\dfrac{1}{7}$.
Therefore, the correct option is (c) and $\dfrac{1}{7}$ is the probability of a leap year having 5 Sundays.


Note: Make sure to remember that only in a leap year the February month has 29 days which means 4 weeks and 1 day extra otherwise it has only 28 days in a month. The value of Probability is always between 0 and 1. If the value of probability is 0 meaning an impossible event and 1 meaning a certain event.