Question

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

Answer 1

Step 1: Determine the possible outcomes

We know that, the total number of days in a leap year $=366$

Since there are complete $52\mathrm{weeks}=364\mathrm{days}$ in a year,

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

Now, since there are $366-364=2$ extra days in a leap year.

Then, there are only two possibilities that one of these two days is a Friday.

i.e., (Thursday, Friday) and (Friday, Saturday)

Step 2: Apply the formula of probability

Here, it is clear that,

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

$\Rightarrow$ Total possible outcomes $=7$

And, the number of favourable outcomes $=2$

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

$\Rightarrow$ Probability of getting $53$ Fridays in a leap year $=\frac{2}{7}$

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