MHT CET · Maths · Probability
The probability, that a year selected at random will have 53 Mondays, is
- A \(\frac{1}{4}\)
- B \(\frac{3}{28}\)
- C \(\frac{5}{28}\)
- D \(\frac{3}{4}\)
Answer & Solution
Correct Answer
(C) \(\frac{5}{28}\)
Step-by-step Solution
Detailed explanation
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.
\(\begin{aligned} \therefore \quad & 53^{\text {rd }} \text { Monday's chance in a leap year }=\frac{2}{7} \\ & \text { Required probability } \\ & =P(\text { a non-leap year and Monday) } \\ & \quad+P(\text { a leap year and Monday }) ~ \\ & =\frac{3}{4} \times \frac{1}{7}+\frac{1}{4} \times \frac{2}{7} \\ & =\frac{5}{28}\end{aligned}\)
\(\therefore \quad\) 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.
\(\begin{aligned} \therefore \quad & 53^{\text {rd }} \text { Monday's chance in a leap year }=\frac{2}{7} \\ & \text { Required probability } \\ & =P(\text { a non-leap year and Monday) } \\ & \quad+P(\text { a leap year and Monday }) ~ \\ & =\frac{3}{4} \times \frac{1}{7}+\frac{1}{4} \times \frac{2}{7} \\ & =\frac{5}{28}\end{aligned}\)
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