A number is chosen randomly from one of the two sets X={2001, 2002, 2003,…., 2100}, Y={1901, 1902, 1903,……, 2000}. If the number chosen represents a calander year and p is the probability that selected year has 53 Sunday, then 2800p is equal to

Answer

To solve the problem, we need to find the probability \( p \) that a randomly chosen year from the sets \( X \) and \( Y \) has 53 Sundays. We will then calculate \( 2800p \).

### Step 1: Identify the sets and years
The sets are:
- \( X = \{2001, 2002, 2003, \ldots, 2100\} \)
- \( Y = \{1901, 1902, 1903, \ldots, 2000\} \)

### Step 2: Count the total number of years in each set
- Set \( X \) has 100 years (from 2001 to 2100).
- Set \( Y \) has 100 years (from 1901 to 2000).

### Step 3: Determine leap years in each set
A leap year is defined as a year that is divisible by 4, except for end-of-century years which must be divisible by 400.

**For Set \( X \):**
- Leap years from 2001 to 2100 are: 2004, 2008, 2012, ..., 2096.
- This gives us the sequence \( 2004 + 4n \) where \( n = 0, 1, 2, \ldots, 24 \).
- Total leap years in set \( X \) = 24.

**For Set \( Y \):**
- Leap years from 1901 to 2000 are: 1904, 1908, 1912, ..., 2000.
- This gives us the sequence \( 1904 + 4n \) where \( n = 0, 1, 2, \ldots, 24 \).
- Total leap years in set \( Y \) = 25.

### Step 4: Calculate the probabilities of having 53 Sundays
**In a non-leap year:**
- A non-leap year has 365 days, which means 52 weeks and 1 extra day. The extra day can be any day of the week.
- Probability of having 53 Sundays = \( \frac{1}{7} \).

**In a leap year:**
- A leap year has 366 days, which means 52 weeks and 2 extra days. The two extra days can be any combination of two consecutive days.
- Probability of having 53 Sundays = \( \frac{2}{7} \).

### Step 5: Calculate the total probability \( p \)
The total probability \( p \) can be calculated as follows:

1. Probability of selecting set \( X \):
- Probability of selecting a leap year from \( X \): \( \frac{24}{100} \)
- Probability of selecting a non-leap year from \( X \): \( \frac{76}{100} \)

Thus, the contribution from set \( X \):
\[ \text{Contribution from } X = \frac{1}{2} \left( \frac{24}{100} \cdot \frac{2}{7} + \frac{76}{100} \cdot \frac{1}{7} \right) \]

2. Probability of selecting set \( Y \):
- Probability of selecting a leap year from \( Y \): \( \frac{25}{100} \)
- Probability of selecting a non-leap year from \( Y \): \( \frac{75}{100} \)

Thus, the contribution from set \( Y \):
\[ \text{Contribution from } Y = \frac{1}{2} \left( \frac{25}{100} \cdot \frac{2}{7} + \frac{75}{100} \cdot \frac{1}{7} \right) \]

3. Combine contributions:
\[ p = \frac{1}{2} \left( \frac{24 \cdot 2 + 76 \cdot 1}{700} + \frac{25 \cdot 2 + 75 \cdot 1}{700} \right) \]
\[ = \frac{1}{2} \left( \frac{48 + 76 + 50 + 75}{700} \right) = \frac{1}{2} \left( \frac{249}{700} \right) = \frac{249}{1400} \]

### Step 6: Calculate \( 2800p \)
\[ 2800p = 2800 \cdot \frac{249}{1400} = 498 \]

### Final Answer
The value of \( 2800p \) is \( 498 \).

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