Answer
P(two or three or four) $=\frac{3}{13}\approx0.231$
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.
Let G be the event "a four". So, as shown in Figure 9, N(G) = 4
The events "a two", "a three" and "a four" are mutually exclusive. So:
P(two or three or four) = P(two) + P(three) + P(four) =
$\frac{N(two)}{N(S)}+\frac{N(three)}{N(S)}+\frac{N(four)}{N(S)}=\frac{4}{52}+\frac{4}{52}+\frac{4}{52}=\frac{12}{52}=\frac{3}{13}\approx0.231$
