1. ## Fish

A pond contains 30 fish. 10 of them are captured and tagged. Once all fish have been tagged they are put back into the pond. The next day 10 of the 30 fish are captured. What is the probability that none of the fish caught is tagged?

Here is what I did:

Let A = {not one of the 10 fish captured is tagged}

$|A| = \binom{20}{10} \times \binom{10}{0} = 184756$ since we are choosing 10 untagged fish from 20 and 0 tagged fish from 10.

$|\Omega| = \binom{30}{10} = 30045015$

$P(A) = \frac{184756}{30045015} = 6.149 \times 10^{-3}$

Is this correct or do I need to include the first part where the fish are initially tagged? Thanks for any help.

2. Originally Posted by slevvio
A pond contains 30 fish. 10 of them are captured and tagged. Once all fish have been tagged they are put back into the pond. The next day 10 of the 30 fish are captured. What is the probability that none of the fish caught is tagged?

Here is what I did:

Let A = {not one of the 10 fish captured is tagged}

$|A| = \binom{20}{10} \times \binom{10}{0} = 184756$ since we are choosing 10 untagged fish from 20 and 0 tagged fish from 10.

$|\Omega| = \binom{30}{10} = 30045015$

$P(A) = \frac{184756}{30045015} = 6.149 \times 10^{-3}$

Is this correct [snip] Mr F says: Yes.
..

3. OK thank you for the help, I thought maybe I had to combine that with all the combinations of fish-tagging but I guess not