Answer
$P(survived~|~child)=\frac{57}{109}\approx0.5229$
Work Step by Step
- First, we need to find $P(child)$:
1) a single event
2) relative frequency
Use the Empirical Approach (page 258):
The sample space: 2224 passengers. So, $N(S)=2224$
According to the marginal distribution (see page 235) of the third column: $N(child)=109$
$P(child)=\frac{N(child)}{N(S)}=\frac{109}{2224}$
- For $P(survived~and~child)$ use relative frequency: Empirical Approach (page 258):
According to the cell in the first row, third column: $N(survived~and~child)=57$
$P(survived~and~child)=\frac{N(survived~and~child)}{N(S)}=\frac{57}{2224}$
- Now, use the Conditional Probability Rule (page 288):
$P(survived~|~child)=\frac{P(survived~and~child)}{P(child)}=\frac{\frac{57}{2224}}{\frac{109}{2224}}=\frac{57}{109}\approx0.5229$
