What is the probability that a leap year selected at random contains 53 Sunday?

Options

A. `2/7`
B. `3/7`
C. `4/7`
D. `5/7`

Explanation

To find the probability that a leap year selected at random contains 53 Sundays, we can follow these steps:

### Step 1: Understand the structure of a leap year
A leap year has 366 days. This can be divided into weeks:
- **Weeks in a leap year**: 366 days ÷ 7 days/week = 52 weeks with a remainder of 2 days.

### Step 2: Identify the number of Sundays
From the 52 complete weeks, there are:
- **Sundays from complete weeks**: 52 Sundays.

### Step 3: Analyze the remaining days
After accounting for the 52 weeks, we have 2 extra days in a leap year. These 2 days can be any combination of the 7 days of the week. The possible combinations of the 2 extra days are:
1. Sunday and Monday
2. Monday and Tuesday
3. Tuesday and Wednesday
4. Wednesday and Thursday
5. Thursday and Friday
6. Friday and Saturday
7. Saturday and Sunday

### Step 4: Count favorable outcomes
To have 53 Sundays in a leap year, one of the 2 extra days must be a Sunday. The combinations that include Sunday are:
- Sunday and Monday
- Saturday and Sunday

Thus, there are **2 favorable outcomes** (Sunday and Monday, Saturday and Sunday).

### Step 5: Calculate total outcomes
From the combinations listed, there are a total of **7 possible combinations** of the 2 extra days.

### Step 6: Calculate the probability
The probability \( P \) of having 53 Sundays in a leap year is given by the formula:
\[ P(\text{53 Sundays}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{2}{7} \]

### Final Answer
The probability that a leap year selected at random contains 53 Sundays is \( \frac{2}{7} \).

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