Question

Find the probability of getting $53$ Sundays in a non-leap year.

• A. $\frac{2}{7}$
• B. $\frac{1}{7}$
• C. $\frac{1}{5}$
• D. $\frac{1}{9}$

Answer 1

The correct option is B

$\frac{1}{7}$

Step 1: Determine the possible outcomes

We know that, the total number of days in a non-leap year $=365$

Since there are complete $52\mathrm{weeks}=364\mathrm{days}$ in a non-leap year.

So, the occurrence of $52$ Sundays in a year is confirmed.

Now, there is $365-364=1$ extra day in a non-leap year.

As this one day can be any of Sunday, Monday, Tuesday, Wednesday, Thursday, Friday or Saturday.

Then, there is only one possibility that this extra day is a Sunday.

Step 2: Apply the formula of probability

Here, it is clear that,

Total possible outcomes $=$Sunday, Monday, Tuesday, Wednesday, Thursday, Friday and Saturday.

$\Rightarrow$ Total possible outcomes $=7$

And, the number of favourable outcomes $=1$

$\because$ Probability of an event $\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{favourable}\mathbf{}\mathbf{outcomes}}{\mathbf{total}\mathbf{}\mathbf{possible}\mathbf{}\mathbf{outcomes}}$

$\Rightarrow$ Probability of getting $53$ Sundays in a leap year $=\frac{1}{7}$

Hence, the probability of getting $53$ Sundays in a leap year is $\frac{1}{7}$.