# A problem in Probability

I have a doubt in a sum of probability. The sum states :

"There are 2 girls. One is born in the year 1989 and the other in 1990. What is the probability that they both have the same birthdays?"

Can anyone please explain me the answer?

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What are your thoughts so far? – Brett Frankel Jan 10 '13 at 15:09
1990 is a step year? – Steve Jan 10 '13 at 15:10
I think it should be 1/365 ...... because there are 2 years and two same days ...... – Pranit Bauva Jan 10 '13 at 15:10
looks like that's correct!! – Santosh Linkha Jan 10 '13 at 15:11
@experimentX I need exact answer because tomorrow is my exam !! – Pranit Bauva Jan 10 '13 at 15:12

Of course, we have to assume something to answer the question. I believe it is indented that the probability of birthday is uniformly distributed over the days of the year. Another thing to check is if one of the years is a leap year. This is not the case for the years 1989 and 1990. So in total there are $365^2$ possibilities for the two birthdays of which $365\cdot 1$ result in an equal birthday (the first girl can have any birthday, but the second has to have exactly the same). So in total we have the probability $$P = \frac{365}{365^2} = \frac{1}{365}.$$
If both years where leap years (e.g., 1988 and 1992) then we would correspondingly have $$P = \frac{1}{366}.$$
A little more difficult is the question when one of the years is a leap year and one not (e.g., 1988 and 1989). Then there are $365\cdot 366$ possibilities of which $365\cdot 1$ result in an equal birthday (there are 365 possibilities for the girl born in the year without leap day and the other girl need to be born at the same day), so $$P= \frac{365}{366\cdot 365} = \frac{1}{366};$$ that is the same result as in the case where both girls are born in a leap year.