Answer
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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.
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.
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