Answer
P(two or three) $=\frac{2}{13}=0.154$
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 three". So, as shown in Figure 9, N(F) = 4
The events "a two" and "a three" are mutually exclusive. So:
P(two or three) = P(two) + P(three) =
$\frac{N(two)}{N(S)}+\frac{N(three)}{N(S)}=\frac{4}{52}+\frac{4}{52}=\frac{8}{52}=\frac{2}{13}=0.154$