Double or combined propbability

#1
Hi I am massively out of my depth here but wanted to know how you would work out something. I am looking at tennis statistics, 1st and second serve.

if i have a player who on first serve loses 40% of them and on second serve loses 20% of them. what is the overall probability of them losing a serve?

I hope this makes sense but cant work out wether they can or cant be merged into a single % loss figure?

If you need more info please ask and I will try to give it you.
 

BGM

TS Contributor
#2
From the tennis rule, a player have a second serve only when he lose the first serve. So I will interpret the figure 20% as a conditional probability, conditional on the first serve already lost.

Then a player losing a serve if and only if he lose both first and second serve. The probability will be just multiplying 40% with 20%:
40% x 20% = 8%

This is a straightforward definition of the conditional probability P(A ⋂ B) = P(A)P(B | A)
 

Dason

Ambassador to the humans
#4
Also this just seems like a generic probability homework kind of question but it'd be interesting to see if there is some sort of interaction between the first/second serve probabilities based on how far ahead/behind in the game the player is.