Question

What is the probability that an ordinary year has $53$ Sundays?

Answer 1

Step 1: Find the number of favorable outcomes:

The number of days in an ordinary year $=365$

The number of weeks in an ordinary year will be

$$\begin{gathered} =\frac{365}{7} \\ =52\text{weeks and 1 day} \end{gathered}$$

Thus, the possible outcomes for that particular day will be $\text{{Mon,Tue,Wed,Thu,Fri,Sat,Sun}}$

The number of possible outcomes for that particular day will be $\text{7}$

Also, the number of favorable outcomes of being a Sunday in that ordinary year is $1$

Step 2: Use the formula for probability:

$\text{Probability=}\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

Thus, the probability will be,

$\text{Probability=}\frac{1}{7}$

Hence, the probability that an ordinary year has $53$ Sundays will be $\frac{1}{7}$