The short answer:
R & J have the same birthday if R's birthday lies on J's birthday, which is one day amoung 365. As a consequence, this happens with probability .
R & J have different birthdays if R's birthday is one the 364 J's non-birthdays, hence this has probability
The long answer:
There are two implicit assumptions in this problem.
- First, the birthdays of Jeena and Reena are independent (for instance, there're not twins ! ), which means that knowing one would not help you guessing what the other one is. This is typically the case indeed.
- Then, we assume that any day has same probability to be Jeena's birthday (or Reena's). This means that if you don't know Jeena at all, and you have to guess what her birthday is, then no day is a better choice than another. This is not quite true (I guess it fluctuates with the seasons, the holidays...), but not that false, so let's forget about these fluctuations. In others words, Jeena's birthday (or Reena's birthday) is any given day with probability .
Now, you want to know if Reena and Jeena have the same birthday, so you need to compare them. Suppose you know Reena's birthday. Because of the independence, knowing R's birthday doesn't affect the fact that any day has same probability to be J's birthday. So the probability that J's birthday lies on R's birthday has probability (you know R's birthday, so it is now like a "fixed" day). This result is the same, whatever day Reena told you she was born: your knowledge of her birthday is of no impact on the probability that the birthdays are the same, so it is just like you did not know anything. As a conclusion, the probability that Reena and Jeena (two random girls you know nothing about) have the same birthday is equal to .
As for your second question, you ask for the probability of the contrary of the event the first question was about. So its probability is one minus the previous probability: the probability that J and R have different birthdays is .
(or you can reproduce the previous reasoning, except that we look at the possibility that J's birthdays lies on any day except R's birthday, which makes 364 possible days, each with probability . )
I hope my quite lengthy explanation has convinced you, if the short one hadn't... I tried to make it as rigorous as possible, underlining the usually hidden arguments.