Answer
$P(both~survived)=\frac{99,540}{180,200}\approx0.5524$
Work Step by Step
- First female:
A single event and the outcomes are equally likely.
The sample space are the 425 females. So, $N(S_1)=425$
316 females survived. Now, consider the event "first female survived". $N(first~female~survived)=316$
Using the Classical Method (page 259):
$P(first~female~survived)=\frac{N(first~female~survived)}{N(S_1)}=\frac{316}{425}$
- Second female:
A single event and the outcomes are equally likely.
The sample space are the 424 remaining females. So, $N(S_2)=424$
There are 315 remaining surviving females . Now, consider the event "second female survived". $N(second~female~survived~|~first~female~survived)=315$
Using the Classical Method (page 259):
$P(second~female~survived~|~first~female~survived)=\frac{N(second~female~survived~|~first~female~survived)}{N(S_2)}=\frac{315}{424}$
- Finally:
$P(both~survived)=P(first~female~survived~and~second~female~survived)$
1) not a single event
2) AND
3) the events are not independent
Use the General Multiplication Rule (page 289):
$P(first~female~survived~and~second~female~survived)=P(first~female~survived)\times P(second~female~survived~|~first~female~survived)=\frac{316}{425}\times\frac{315}{424}=\frac{99,540}{180,200}\approx0.5524$
