Question

The probability that a leap year will have 53 Fridays or 53 Saturdays is
(a) 2/7
(b) 3/7
(c) 4/7
(d) 1/7

Answer 1

(b) 3/7

We know that a leap year has 366 days (i.e. 7 $\times$ 52 + 2) = 52 weeks and 2 extra days
The sample space for these 2 extra days is given below:
S = {(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)}
There are 7 cases.
∴ n(S) = 7
Let E be the event that the leap year has 53 Fridays or 53 Saturdays.
E = { (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)}
i.e. n(E) = 3
$\therefore P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{3}{7}$∴P(E)=n(E)n(S)=37
Hence, the probability that a leap year has 53 Fridays or 53 Saturdays is $\frac{3}{7}$.