The probability that a leap year selected ar random contains either 53 sundays or 53 mondays, is

Options

A. `(1)/(7)`
B. `(2)/(7)`
C. `(3)/(7)`
D. `(4)/(7)`

Answer

To solve the problem of finding the probability that a leap year selected at random contains either 53 Sundays or 53 Mondays, we can follow these steps:

### Step 1: Understand the total number of days in a leap year
A leap year has 366 days. This means there are 52 complete weeks (which account for 364 days) and 2 extra days.

**Hint:** Remember that a week has 7 days, so 52 weeks will cover 364 days, leaving 2 extra days.

### Step 2: Identify the possible combinations of the extra days
The two extra days can be any of the following combinations:
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

**Hint:** List all possible pairs of consecutive days to see which days can be extra in a leap year.

### Step 3: Determine when there are 53 Sundays or 53 Mondays
From the combinations of extra days:
- There will be 53 Sundays if the extra days are:
- Sunday and Monday
- Saturday and Sunday

- There will be 53 Mondays if the extra days are:
- Sunday and Monday
- Monday and Tuesday

**Hint:** Check which combinations include Sunday or Monday as one of the extra days.

### Step 4: Count the favorable outcomes
- The combinations that give us 53 Sundays are:
1. Sunday and Monday
2. Saturday and Sunday

- The combinations that give us 53 Mondays are:
1. Sunday and Monday
2. Monday and Tuesday

Now, we can see that:
- The combination "Sunday and Monday" is counted in both cases.

Thus, we have:
- Total favorable outcomes for 53 Sundays = 2
- Total favorable outcomes for 53 Mondays = 2
- The overlap (both Sundays and Mondays) = 1

### Step 5: Use the principle of inclusion-exclusion
To find the total number of favorable outcomes (either 53 Sundays or 53 Mondays), we can use the formula:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

Where:
- \( P(A) \) = Probability of 53 Sundays = 2/7
- \( P(B) \) = Probability of 53 Mondays = 2/7
- \( P(A \cap B) \) = Probability of both 53 Sundays and 53 Mondays = 1/7

Substituting these values:
\[ P(A \cup B) = \frac{2}{7} + \frac{2}{7} - \frac{1}{7} = \frac{3}{7} \]

### Step 6: Conclusion
The probability that a leap year selected at random contains either 53 Sundays or 53 Mondays is \( \frac{3}{7} \).

**Final Answer:** The probability is \( \frac{3}{7} \).

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