Calculate the probability of two teams have been drawns If we know that team A had a $39\%$ chance of winning and team B $43\%$ chance of winning, how we can calculate the probability of the teams drawn?
My textbook mention the answer but I cannot understand the logic behind it. The answer is $18\%$. As working is not shown I guess that this is how the find $18\%$ probability of two teams withdrawn:
$$ (100\% - 39\%) - 43\% = 18\%$$
But I cannot understand the logic behind it. I appreciate if someone can explain it to me.

The sum of all events' probabilities is equal to 1. In this case, there are three disjoint events: team A winning, team B winning or a draw. Since we know the sum of these probabilities is 1, we can get the probability of a draw as follows:
$$
Pr(\text{Draw})=1-Pr(\text{Team A wins})-Pr(\text{Team B wins})=1-0.39-0.43=0.18
$$
