• Sep 28th 2010, 07:37 AM
Runty
I need a confirmation on the answer I have here.

A small lake has 105 fish: 40 are trout and 65 are carp. A fisherman caught 8 fish. What is the probability that exactly 2 of them are trout if we know that at least 3 of them are not.

My current answer is as follows:
$P=\frac{C(65,3)\times C(40,5)}{C(105,8)}+\frac{C(65,4)\times C(40,4)}{C(105,8)}+\frac{C(65,5)\times C(40,3)}{C(105,8)}+\frac{C(65,6)\times C(40,2)}{C(105,8)}$

If I did anything wrong, can someone point it out?
• Sep 28th 2010, 07:52 AM
Quote:

Originally Posted by Runty
I need a confirmation on the answer I have here.

A small lake has 105 fish: 40 are trout and 65 are carp. A fisherman caught 8 fish. What is the probability that exactly 2 of them are trout if we know that at least 3 of them are not.

My current answer is as follows:
$P=\frac{C(65,3)\times C(40,5)}{C(105,8)}+\frac{C(65,4)\times C(40,4)}{C(105,8)}+\frac{C(65,5)\times C(40,3)}{C(105,8)}+\frac{C(65,6)\times C(40,2)}{C(105,8)}$

If I did anything wrong, can someone point it out?

You are calculating the probability of the fisherman catching "at least 2" trout and "at least" 3 carp..

That's a different question.

The question is...

The fisherman caught 8 fish.
3 of them are definately carp.
What's the probability that of the remaining 5 fish caught, exactly 2 are trout.
• Sep 28th 2010, 07:57 AM
Runty
Darn it, wrong way.

Think you might be able to show a correction or at least help me with a new start-up?

EDIT: Out of curiosity, would this be considered a conditional probability question?
• Sep 28th 2010, 04:19 PM
mr fantastic
Quote:

Originally Posted by Runty
Darn it, wrong way.

Think you might be able to show a correction or at least help me with a new start-up?

EDIT: Out of curiosity, would this be considered a conditional probability question?

Yes, it is a conditional probability question.
• Sep 29th 2010, 05:03 AM
Quote:

Originally Posted by Runty
Darn it, wrong way.

Think you might be able to show a correction or at least help me with a new start-up?

EDIT: Out of curiosity, would this be considered a conditional probability question?

Hi Runty,

if we denote A as the event that the fisherman caught 6 carp and 2 trout,
while denoting B to be the event of at least 3 of the 8 fish being carp (think of it as picking 3 fish from the 8 caught and they happen to be carp
hence we know that at least 3 are carp),

then we need to calculate the probability that the fisherman caught 6 carp given that "a random selection" of 3 of the 8 all happen to be carp.

This is

$\displaystyle\frac{P(fisherman\ caught\ 6\ carp\ and\ 2\ trout)}{P(at\ least\ 3\ carp\ were\ caught)}$

$\displaystyle\Pr( A | B )=\frac{P(AB)}{P(B)}$

The event AB of 3 carp being caught and 6 carp being caught reduces to 6 carp caught.

Therefore you need to calculate the number of ways of catching 6 carp and 2 trout for the numerator.
Calculate the numerator probability.

The denominator is the probability of catching at least 3 carp.

You'll have to sum the probabilities of catching 3, 4, 5, 6, 7 or 8 carp,
or sum the probabilities of catching 0, 1 or 2 carp and subtract from 1.

See how you get on with that.
• Oct 1st 2010, 10:22 AM
The fisherman can catch 6 carp and 2 trout in $\binom{65}{6}\binom{40}{2}$ ways,

hence the probability of catching 6 carp and 2 trout is $\displaystyle\frac{\binom{65}{6}\binom{40}{2}}{\bi nom{105}{8}}$

The probability of catching "less" than 3 carp is obtained by summing the probabilities of
catching 0, 1 or 2 carp.

Then the probability of catching at least 3 carp is 1-P(<3\ carp)

The probability of catching 0 carp is the probability of catching 8 trout, which is $\displaystyle\frac{\binom{65}{0}\binom{40}{8}}{\bi nom{105}{8}}$

The probability of catching 1 carp is the probability of catching 1 carp and 7 trout, which is $\displaystyle\frac{\binom{65}{1}\binom{40}{7}}{\bi nom{105}{8}}$

The probability of catching 2 carp and 6 trout is $\displaystyle\frac{\binom{65}{2}\binom{40}{6}}{\bi nom{105}{8}}$

Therefore, the probability of catching exactly 2 trout given that at least 3 of the 8 fish caught are carp is

$\displaystyle\frac{\frac{\binom{65}{6}\binom{40}{2 }}{\binom{105}{8}}}{1-\frac{\binom{65}{0}\binom{40}{8}+\binom{65}{1}\bin om{40}{7}+\binom{65}{2}\binom{40}{6}}{\binom{105}{ 8}}}$

$=0.239$ approx.