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 As working is not shown I guess that this is how the find $18\%$ probability of two teams withdrawn: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.

Let $A$ be the event that  team A wins, let $B$ be the event team B wins, and let $D$ be the event they draw. Then since precisely one of these events must happen,
$$\Pr(A)+\Pr(B)+\Pr(D)=1.$$
