The probability that a non leap year selected at random will have 53 Sundays is
A.$0$
B.$\dfrac{1}{7}$
C.$\dfrac{2}{7}$
D.$\dfrac{3}{7}$

Hint : (Consider all the possibilities of a non-leap year)

A non-leap year has $365$days i.e. $52$ weeks and $1$ odd days.

So, there are $52$ Sundays always.

But the odd day can be Sunday, Monday, Tuesday, Wednesday, Thursday, Friday or Saturday.

So, there are $7$possibilities for odd days and out of which one is favourable.

So, the probability of getting Sunday on last week is $\dfrac{1}{7}$​

Thus, the probability of getting$\,\,53\,\,$Sundays in a non-leap year is $\dfrac{1}{7}$​.
So the correct option is B.

Note :- In these types of questions of probability rather than thinking of complex situations, we have to think creatively. Here in this question we have considered all the probabilities of a non-leap year then according to that we have solved the question.