Question

Find the probability of getting $53$ Fridays in a leap year.

Answer 1

Step 1: Find the number of favourable outcomes:

A leap year consists of 366 days.

The number of weeks in a year will be 52.

The number of weeks in a leap year will be,

$$\begin{gathered} =\frac{366}{7} \\ =52weeksand2days. \end{gathered}$$

Thus, the number of possible outcomes for the two days will be,

$\text{=(Mon,Tue),(Tue,Wed),(Wed,Thurs),(Thurs,Fri),(Fri,Sat),(Sat,Sun),(Sun,Mon)}$

The total number of possible outcomes will be $7$.

Thus, the favourable outcomes being a Friday will be $\text{(Thurs,Fri),(Fri,Sat)}$

Therefore, the number of favourable outcomes will be $2$

Step 2: Use the formula for probability:

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

Thus, the probability will be,

$\text{Probability=}\frac{2}{7}$

Final answer:

Hence, the probability of getting $53$ Fridays in a leap year is $\frac{2}{7}$.