Question

Find the probability of having 53 Saturdays in:
i) a non - leap year
ii) a leap year
[4 MARKS]

Answer 1

Each subpart: 2 Marks each

(i) In a non-leap year, there are 365 days and 364 days make 52 weeks. In a week, there is one Saturday.
It means we have to find the probability of having a Saturday out of the remaining 1 day.
$\therefore$ Probability (having 53 Saturdays)
$=\frac{n\left(E\right)}{n\left(S\right)}=\frac{1}{7}$

ii) We know that in a leap year, there are 366 days and 364 days make 52 weeks. In each week, there is one Saturday. It means we have to find the probability of having a Saturday out of the remaining 2 days.
The remaining two days can either be a (Sunday-Monday),(Monday-Tuesday),(Tuesday-Wednesday),(Wednesday-Thursday),(Thursday-Friday),(Friday-Saturday),(Saturday-Sunday)
$\therefore$ Probability(having 53 Saturdays)
$=\frac{n\left(E\right)}{n\left(S\right)}=\frac{2}{7}$