(1) Both of them can have their birthdays only in one of the seven days of a week. So, the probability that they have birthday on the same day is 1/7

Problem: Find the probability that Savita and Hamida will have the same birthday and different birthdays.

Assumptions:
- We assume that there are 365 days in a year and all birthdays are equally likely.
- We ignore the possibility of a leap year.

Solution:

Same Birthday:
- To find the probability that Savita and Hamida will have the same birthday, we need to consider the number of favorable outcomes and the total number of possible outcomes.
- The favorable outcomes occur when both Savita and Hamida have the same birthday, i.e., there is only one specific day out of the 365 days.
- The total number of possible outcomes is 365 * 365 since each person can have a birthday on any of the 365 days.
- Therefore, the probability of having the same birthday is 1 / (365 * 365) ≈ 0.0000767 or 0.00767%.

Different Birthdays:
- To find the probability that Savita and Hamida will have different birthdays, we need to consider the number of favorable outcomes and the total number of possible outcomes.
- The favorable outcomes occur when Savita and Hamida have different birthdays, i.e., there are 365 * 364 different combinations.
- The total number of possible outcomes is still 365 * 365 since each person can have a birthday on any of the 365 days.
- Therefore, the probability of having different birthdays is (365 * 364) / (365 * 365) ≈ 0.99726 or 99.726%.

Conclusion:
- The probability that Savita and Hamida will have the same birthday is approximately 0.00767%.
- The probability that Savita and Hamida will have different birthdays is approximately 99.726%.
- These probabilities assume that all birthdays are equally likely and ignore the possibility of a leap year.