Answer
P(two or club) $=\frac{4}{13}\approx0.308$
Work Step by Step
The sample space S = "a standard 52-card deck". So, N(S) = 52.
Let E be the event "a two". So, as shown in Figure 9, N(E) = 4.
Let F be the event "a club card". So, as shown in Figure 9, N(F) = 13.
The events "a two" and "a club card" are not mutually exclusive. There is one outcome in common, the two of clubs, which means N(two and club) = 1. Now, using The General Addition Rule:
P(two or club) = P(two) + P(club) - P(two and club) =
$\frac{N(two)}{N(S)}+\frac{N(three)}{N(S)}-\frac{N(\text{two and club})}{N(S)}=\frac{4}{52}+\frac{13}{52}-\frac{1}{52}=\frac{16}{52}=\frac{4}{13}\approx0.308$
