Answer
$P(both~chips~are ~defective)=\frac{49}{1999800}\approx0.00002450$
Work Step by Step
- First chip:
The sample space are the 10,000 chips. So, $N(S_1)=10,000$
There are 50 defective chips. Now, consider the event "first chip is defective". $N()=50$
Using the Classical Method (page 259):
$P(first~chip~is~defective)=\frac{N(first~chip~is~defective)}{N(S_1)}=\frac{50}{10,000}=\frac{1}{200}$
- Second chip:
The sample space are the 9999 remaining chips. So, $N(S_2)=9999$
There are 49 remaining defective chips. Now, consider the event "second chip is defective". $N(second~chip~is~defective~|~first~chip~is~defective)=49$
Using the Classical Method (page 259):
$P(second~chip~is~defective~|~first~chip~is~defective)=\frac{N(second~chip~is~defective~|~first~chip~is~defective)}{N(S_2)}=\frac{49}{9999}$
Now, using the General Multiplication Rule (page 289):
$P(both~chips~are ~defective)=P(first~chip~is~defective)\times P(second~chip~is~defective~|~first~chip~is~defective)=\frac{1}{200}\times\frac{49}{9999}=\frac{49}{1999800}\approx0.00002450$
