Answer
$P(female~or~a~child)=\frac{534}{2224}\approx0.2401$
Work Step by Step
The sample space: 2224 passengers. So, $N(S)=2224$
First, let's find $P(child)$:
It is a single event and we will use relative frequency (Empirical Approach, page 258):
According to the marginal distribution (see page 235) of the third column: $N(child)=109$
$P(child)=\frac{N(child)}{N(S)}=\frac{109}{2224}$
From (b): $P(female)=\frac{425}{2224}$
Now, let's find $P(female~or~a~child)$:
1) OR
2) mutually exclusive (disjoint events)
Use the Addition Rule for Disjoint Events (page 270):
$P(female~or~a~child)=P(female)+P(child)=\frac{425}{2224}+\frac{109}{2224}=\frac{534}{2224}\approx0.2401$
