There are 5 girls and 5 boys in a chess club. the club holds a round-robin tournament in which every player plays against every other player exactly once.

in how many games does a girl play against a boy? ( 25)

in how many games does a girl play against another girl? ( 10)

what fraction of the games are boy-versus-boy? enter your answer as a fraction in simplified form.( 2/9)

if anita is one of the club members, what fraction of all games at the tournament does anita play in? enter your answer as a fraction in simplified form.( i have no idea)
(this single question worth 64 points)

1. In how many games does a girl play against a boy?
A: Because there are 5 girls, each will play against 5 boys, so in total is 5*5 = 25 times.

2. In how many games does a girl play against another girl?
A: In this case, we only need to calculate how many girl pairs are there among these 5 girls, and it become a combination problem which can solved by 

3.The fraction of games "B vs B" should be the number of "B vs B" (which is 10) divide the total number of the games "B vs B" & "G vs G" & "B vs G" (which is 10+10+25=45, see the analysis in 1.A and 2.A), so the fraction should be 10/45=2/9

4.One person will have to play against 9 people at the tournament, and the total number of the games is 45, so the fraction should be 9/45 = 1/5

This does not make sense

Um what? rephrase that ?