Probability — Maths STD 11 Science — Question

$\therefore$ The probability of a year being a leap year $=\frac{1}{4}$

$\therefore$ Probability of a year being a non-leap year $=1-\frac{1}{4}=\frac{3}{4}$

In a non-leap year, there are 52 weeks and one extra day, whereas a leap year has 52 weeks and 2 extra days.

$\therefore 53$ rd Wednesday's chance in a non-leap year $=\frac{1}{7}$

Two extra days of a leap year can be (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun), (Sun, Mon)

∴ There are 2 possibilities of 53rd Wednesday in a leap year.

$\therefore 53$ rd Wednesday's chance in a leap year $=\frac{2}{7}$

Required probability = P(a non-leap year and Wednesday) + P(a leap year and Wednesday)

$\begin{aligned} & =\frac{3}{4} \times \frac{1}{7}+\frac{1}{4} \times \frac{2}{7} \\ & =\frac{5}{28}\end{aligned}$