I'm just looking for a confirmation. Textbook answer for this problem is $\frac{1}{18}$ but I'm getting $\frac{1}{20}$...
You have 7 tickets to a baseball game. Three tickets are in the left field stands, 2 are behind home plate and 2 are near first base. Two tickets are randomly given to your sister. You notice that one of the tickets is not behind home plate. Calculate the probability that both tickets are near first base.
$P(A|B)=\cfrac{\frac{1}{21}}{\frac{20}{21}}=\cfrac{1}{20}$
Am I missing something? Is there a faster way to solve these problems than doing a probability tree or listing the possible outcomes?