anita,sita,gita and rita are 4 friends .What is the probability that(in a non-leap year)

1)all will have same birthday

2) their birthday falls in the month of october.

3) their bday falls on the 10th day of the month

4) their birthday falls in january and february

(1) All will have same birthday
   Probability of Anita's birthday = $\frac{365}{365}$
 Now Sita,Gita and Rita have their birthdays fixed .
 So each of them has the probability  $\frac{1}{365}$
Thus required probability = $(\frac{365}{365}\times \frac{1}{365}\times \frac{1}{365}\times \frac{1}{365})=\left(\frac{1}{{365}^3}\right)$

(2) Their birthdays fall in the month of october.
   As the month of october consists of 31 days.
  In this case ,each of the four has the probability  $\left(\frac{31}{365}\right)$
Thus required probability = $(\frac{31}{365}\times \frac{31}{365}\times \frac{31}{365}\times \frac{31}{365}) = (\frac{31}{365}{)}^4$

(3)Their birthdays fall on the ${10}^{th}$ day of the month
  So in every month there comes the date ${10}^{th}$
  So number of these days = number of months = 12
 So in this case , each of the four has the probability $\frac{12}{365}$
Thus required probability = $(\frac{12}{365}\times \frac{12}{365}\times \frac{12}{365}\times \frac{12}{365}) = (\frac{12}{365}{)}^4$

(4)Their birthdays fall in January and February 
   As January consists of 31 days and February consists of 28 days (non-leap year)
  So total number of days in January and February = (31+28) = 59 days
 So in this case, each of the four has the probability $\frac{59}{365}$
  Thus required probability = $(\frac{59}{365}\times \frac{59}{365}\times \frac{59}{365}\times \frac{59}{365})= (\frac{59}{365}{)}^4$