The probability, that a year selected at random will have 53 Mondays, is

The probability, that a year selected at random will have 53 Mondays, is
  1. $\frac{1}{4}$
  2. $\frac{3}{28}$
  3. $\frac{5}{28}$
  4. $\frac{3}{4}$

Solution

A leap year comes after 3 years. $\therefore \quad$ The probability of a year being a leap year $=\frac{1}{4}$ $\therefore \quad$ 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. $\therefore \quad 53^{\text {rd }}$ Monday'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), (The, Fri), (Fri, Sat), (Sat, Sun), (Sun, Mon) $\therefore \quad$ There are 2 possibilities for having a $53^{\text {rd }}$ Monday in a leap year. $\begin{aligned} \therefore \quad & 53^{\text {rd }} \text { Monday's chance in a leap year }=\frac{2}{7} \\ & \text { Required probability } \\ & =P(\text { a non-leap year and Monday) } \\ & \quad+P(\text { a leap year and Monday }) ~ \\ & =\frac{3}{4} \times \frac{1}{7}+\frac{1}{4} \times \frac{2}{7} \\ & =\frac{5}{28}\end{aligned}$

Asked in: MHT CET 2024 (15 May Shift 1)

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