Question

The probability that a leap year will have 53 Fridays or 53 Saturdays is
(a) $\frac{2}{7}$

(b) $\frac{3}{7}$

(c) $\frac{4}{7}$

(d) $\frac{1}{7}$

Answer 1

$\left(\mathrm{b}\right)\frac{3}{7}$

A leap year has 366 days

For a non-leap year:
52 weeks + 1 day

For a leap year:
52 weeks + 2 days

$$\begin{gathered} \text{Sample space}=\left[\left(\text{Monday, Tuesday}\right),\left(\text{Tuesday, Wednesday}\right),\left(\text{Wednesday, Thursday}\right), \\ \left(\text{Thursday, Friday}\right),\left(\text{Friday, Saturday}\right),\left(\text{Saturday, Sunday}\right),\left(\text{Sunday, Monday}\right)\right] \\ \text{Favourable cases}=3 \\ P\left(\text{53 Fridays or 53 Saturdays}\right)=\frac{3}{7} \end{gathered}$$