Answer:
28% (if you're specifically asking for a win then lose)
46% (if the order doesn't matter and it can be a win/lose or lose/win)
Explanation:
Ok, so the question you provided in the comments is: "chance of Ivan to win one game and lose the other."
I don't know if you mean "what is the probability of winning one game first, and then losing one game" in that order, or in either order.
I'll just calculate the probability of winning one game, and then losing and then give that probability (if that's what you're asking for), and then continue the question if you're asking for either order.
So the probability of winning a game is "70%"
Now the probability of "losing" after that is 40% as given in the graph.
This is only going to occur 40% of the time, in which you win 70% of the time, since you have to actually win first.
So we calculate 40% of 70%, which is just 28%
This is the probability of winning, and then losing, now just take this answer if you're asking this specific order.
If you're asking for either win/lose or lose/win, then we now have to calculate the probability of losing then winning.
The probability losing is 30% initially, as given in the diagram. After that there is a 60% probability of winning. So the chance of losing/winning is going to be 60% of 30%, since you have to lose first, and then win.
So 30% of the time you're going to lose
Then the 30% of the time you're going to lose, 60% of that time you're going to win.
Or in other words 60% of 30% of the time you're going to lose then win.
This simplifies to: 18%
So to calculate the probability of a win/lose or lose/win we just add these probabilities up together
18% + 28% = 46%