Find the probability that the month of February may have 5 Wednesdays in (i) a leap year (ii) a non-leap year.

Answer

To solve the problem of finding the probability that the month of February may have 5 Wednesdays in (i) a leap year and (ii) a non-leap year, we can follow these steps:

### Step-by-Step Solution:

**(i) Probability in a Leap Year:**

1. **Determine the number of days in February during a leap year:**
- In a leap year, February has 29 days.

2. **Calculate the number of complete weeks:**
- There are 7 days in a week.
- To find the number of complete weeks in 29 days, we divide 29 by 7.
- \( 29 \div 7 = 4 \) complete weeks with a remainder of 1 day.
- This means there are 4 complete weeks and 1 extra day.

3. **Count the number of Wednesdays:**
- Each complete week contributes 1 Wednesday, so from the 4 complete weeks, we have 4 Wednesdays.
- The remaining 1 day can be any day of the week.

4. **Determine the probability of the extra day being a Wednesday:**
- The probability that this extra day is a Wednesday is \( \frac{1}{7} \) (since there are 7 days in a week).

5. **Final probability for a leap year:**
- The probability of having 5 Wednesdays in February of a leap year is \( \frac{1}{7} \).

**(ii) Probability in a Non-Leap Year:**

1. **Determine the number of days in February during a non-leap year:**
- In a non-leap year, February has 28 days.

2. **Calculate the number of complete weeks:**
- Again, dividing 28 by 7 gives us \( 28 \div 7 = 4 \) complete weeks with no remainder.
- This means there are 4 complete weeks and no extra days.

3. **Count the number of Wednesdays:**
- From the 4 complete weeks, we have 4 Wednesdays.

4. **Determine the probability of having a fifth Wednesday:**
- Since there are no extra days left, it is impossible to have a 5th Wednesday.
- Thus, the number of favorable outcomes is 0.

5. **Final probability for a non-leap year:**
- The probability of having 5 Wednesdays in February of a non-leap year is \( \frac{0}{7} = 0 \).

### Summary of Results:
- Probability of 5 Wednesdays in a leap year: \( \frac{1}{7} \)
- Probability of 5 Wednesdays in a non-leap year: \( 0 \)