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Q5:
If a leap year is selected at random, what is the chance that it will contain 53 tuesdays?

Solution :

Step 1: Analyzing the Timeframe of a Leap Year

A standard year consists of $365$ days, whereas a leap year consists of $366$ days [Per the Gregorian calendar definition].

We convert the total number of days into weeks:

$366 \text{ days} = 52 \text{ weeks} + 2 \text{ days}$

This implies that every leap year is guaranteed to have $52$ Tuesdays, as $52 \times 7 = 364$ days cover $52$ full cycles of the week.

Step 2: Determining the Sample Space of Remaining Days

The remaining $2$ days (the $365^{th}$ and $366^{th}$ days) can be any two consecutive days of the week. We denote the sample space $S$ as follows:

  • $S = \{(\text{Sunday, Monday}), (\text{Monday, Tuesday}), (\text{Tuesday, Wednesday}), (\text{Wednesday, Thursday}), (\text{Thursday, Friday}), (\text{Friday, Saturday}), (\text{Saturday, Sunday})\}$

The total number of outcomes is $n(S) = 7$.

Step 3: Defining the Favorable Event

For the year to contain exactly $53$ Tuesdays, one of the two remaining days must be a Tuesday. Let $E$ be the event of selecting such a pair:

  • $E = \{(\text{Monday, Tuesday}), (\text{Tuesday, Wednesday})\}$

The number of favorable outcomes is $n(E) = 2$.

Step 4: Calculation of Probability

Using the Classical Definition of Probability [Probability $P(E) = \frac{n(E)}{n(S)}$]:

$P(E) = \frac{2}{7}$

Visual Representation of Sample Space

Sample Space of 2 Remaining Days 1. (Sun, Mon) 2. (Mon, Tue) - Favorable 3. (Tue, Wed) - Favorable 4. (Wed, Thu) 5. (Thu, Fri) 6. (Fri, Sat) 7. (Sat, Sun)

Final Solution: The probability that a leap year contains 53 Tuesdays is $\frac{2}{7}$.


More Questions from Class 12 Mathematics Probability Miscellaneous Exercise on Chapter 13


CBSE Solutions for Class 12 Mathematics Probability


Chapters in CBSE - Class 12 Mathematics


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