MacFauz
A certain business school has 500 students, and the law school at the same university has 800 students. Among these students, there are 30 sibling pairs consisting of 1 business student and 1 law student. If 1 student is selected at random from both schools, what is the probability that a sibling pair is selected?
A. \(\frac{3}{40000}\)
B. \(\frac{3}{20000}\)
C. \(\frac{3}{4000}\)
D. \(\frac{9}{400}\)
E. \(\frac{6}{130}\)
However, I selected B.

Since order doesn't matter, I added \(\frac{3}{40000}\) to \(\frac{3}{40000}\) and got \(\frac{3}{20000}\). Could someone explain why I am wrong??
A certain business school has 500 students, and the law school at the same university has 800 students. Among these students, there are 30 sibling pairs consisting of 1 business student and 1 law student. If 1 student is selected at random from both schools, what is the probability that a sibling pair is selected?
A) 3/40,000 B) 3/20,000 C) 3/4,000 D) 9/400 E) 6/130
I have a question since I choose answer choice B
My reasoning was that we can choose one business school or law school sibling, meaning we can choose business school student first or law school student first: 30/500*1/800= 3/40,000 or 30/800*1/500= 3/40,000 then add up totals 3/40,000+3/40,000=3/20,000.
Because in some questions we do add totals but why don't we add up totals in this question?
Thanks!!!