Can someone help me understand the answer of this question?

[nasi112]

I waited for long for this site to be fixed but it took longer than usual. I've had this question for a while and thought I'd post it now.

In a small lake, it is estimated that there are approximately 105 fish, of which 40 are trout and 65 are carp. A fisherman caught eight fish; what is the probability that exactly two of them are trout if we know that at least three of them are not?

The answer

[imath]\displaystyle\frac{\frac{\binom{40}{2}\binom{65}{6}}{\binom{105}{8}}}{1 - \sum_{i=0}^{2}\frac{\binom{40}{8 - i}\binom{65}{i}}{\binom{105}{8}}} = 0.239[/imath]

I tried to reduce the answer to

[imath]\displaystyle \frac{A}{B} = 0.239[/imath]

Is A the probability 2 trout were caught?
Is B the probability at least 3 carp were caught?

In my understanding, A means the probability the fisherman caught 2 trout AND the fisherman caught 6 carp. 8 fish out of 105.
B could mean the probability of (the fisherman caught 3 carp AND the fisherman caught 5 trout) Or (the fisherman caught 4 carp AND the fisherman caught 4 trout) Or (the fisherman caught 5 carp AND the fisherman caught 3 trout) Or (the fisherman caught 6 carp AND the fisherman caught 2 trout). Also 8 fish out of 105.
B could also mean the probability of (the fisherman caught 3 carp AND the fisherman caught 5 trout) Or (the fisherman caught 4 carp AND the fisherman caught 4 trout) Or (the fisherman caught 5 carp AND the fisherman caught 3 trout) Or (the fisherman caught 6 carp AND the fisherman caught 2 trout) Or (the fisherman caught 7 carp AND the fisherman caught 1 trout) Or (the fisherman caught 8 carp AND the fisherman caught 0 trout). Also 8 fish out of 105.

A is fixed.
B varies.

I think A is very clear.
I am confused about B. Is it possible for the fisherman to catch 7 carp?

Why does the division [imath]\displaystyle\frac{A}{B}[/imath] mean the probability of exactly two of them are trout if we know that at least three of them are not when there's a possibility in B none trout fish were caught.

 

[Dr.Peterson]

The problem is asking for

P(exactly 2 trout | at least 3 carp)​

= P(exactly 2 trout | at most 5 trout).​

This is equal to

P(exactly 2 trout AND at most 5 trout) / P(at most 5 trout)​

= P(exactly 2 trout) / P(at most 5 trout),​

since exactly 2 implies at most 5.

And the numerator and denominator of the given answer agree with this. In particular,

P(at most 5 trout) = 1 - P(more than 5 trout) = 1 - P(less than 3 carp) = 1 - P(0, 1, or 2 carp)​

Does that help?

 

[Agent Smith]

cccttttt
cccctttt
cccccttt
cccccctt
ccccccct
cccccccc

1/6 .. unpack

 

[logistic_guy]

cccttttt
cccctttt
cccccttt
cccccctt
ccccccct
cccccccc

1/6 .. unpack

I like how you crack problems Agent Smith. But I'm sure that you don't know the following:

\(\displaystyle \frac{\binom{40}{2}\binom{65}{6}}{\binom{105}{8}} = \binom{8}{2} \left( \frac{40}{105} \times \frac{39}{104} \times \frac{65}{103} \times \frac{64}{102} \times \frac{63}{101} \times \frac{62}{100} \times \frac{61}{99} \times \frac{60}{98} \right)\)

Or

\(\displaystyle \frac{\binom{40}{2}\binom{65}{6}}{\binom{105}{8}} = \binom{8}{6} \left( \frac{40}{105} \times \frac{39}{104} \times \frac{65}{103} \times \frac{64}{102} \times \frac{63}{101} \times \frac{62}{100} \times \frac{61}{99} \times \frac{60}{98} \right)\)

 

[nasi112]

The problem is asking for

P(exactly 2 trout | at least 3 carp)​

= P(exactly 2 trout | at most 5 trout).​

This is equal to

P(exactly 2 trout AND at most 5 trout) / P(at most 5 trout)​

= P(exactly 2 trout) / P(at most 5 trout),​

since exactly 2 implies at most 5.

And the numerator and denominator of the given answer agree with this. In particular,

P(at most 5 trout) = 1 - P(more than 5 trout) = 1 - P(less than 3 carp) = 1 - P(0, 1, or 2 carp)​

Does that help?

Yes it helps. I got it partially. Thanks doctor.

I understand the conditional if it is written [imath]\displaystyle P(A|B) = \frac{P(A \ AND \ B)}{P(B)}[/imath]

You wrote [imath]\displaystyle \frac{P(A \ AND \ B)}{P(B)} = \frac{P(A)}{P(B)}[/imath].

This part I don't understand fully. I learnt before [imath]\displaystyle P(A \ AND \ B)[/imath] is the intersection between A and B? Am I wrong?

I need to know what is A and what is B to find the intersection.

I will show you example to clarify my idea. If I have X and Y.

X = {A, B, C, 0, 9}
Y = {M,N,C,0,1}

X AND Y = {C,0}

This is my understanding of the intersection. If A AND B in the context of this exercise means something else, I need to understand what it means.

I can understand better if you answer my questions in op post.

cccttttt
cccctttt
cccccttt
cccccctt
ccccccct
cccccccc

1/6 .. unpack

unpack what?

cccttttt means the fisherman caught carp carp carp trout trout trout trout trout
so this is one possibility of what will happen in the denominator

my question was about ccccccct and cccccccc
Can the fisherman catch 7 carp? If yes, then he will catch only 1 trout but the question said exactly 2. This is why I am confused.
The same thing happen for cccccccc means zero trout were caught but what about the condition 2 trout exactly caught.

I understand the order doesn't matter in this case. That's why the writer used combination.

This means cccccctt is one way to catch the fish and ttcccccc another way to catch the fish.

 

[Dr.Peterson]

I understand the conditional if it is written [imath]\displaystyle P(A|B) = \frac{P(A \ AND \ B)}{P(B)}[/imath]

You wrote [imath]\displaystyle \frac{P(A \ AND \ B)}{P(B)} = \frac{P(A)}{P(B)}[/imath].

This part I don't understand fully. I learnt before [imath]\displaystyle P(A \ AND \ B)[/imath] is the intersection between A and B? Am I wrong?

I need to know what is A and what is B to find the intersection.

If A is a subset of B, then A and B = A. Do you see this?
That's why I could say

P(exactly 2 trout AND at most 5 trout) / P(at most 5 trout)= P(exactly 2 trout) / P(at most 5 trout),
since exactly 2 implies at most 5.

Whenever you catch exactly 2 trout, you have caught at most (that is, no more than) 5 trout. So "exactly 2 trout" if a subset of "at most 5 trout".

I will show you example to clarify my idea. If I have X and Y.

X = {A, B, C, 0, 9}
Y = {M,N,C,0,1}

X AND Y = {C,0}

This is my understanding of the intersection. If A AND B in the context of this exercise means something else, I need to understand what it means.

Here is an example of what I said: If X = {A, B, C} and Y = {A, B, C, 0, 1}, then X and Y = {A, B, C} = X.

I can understand better if you answer my questions in op post.

I think I did.

 

[Agent Smith]

There are only trout and carp in the lake.
8 fish were caught
At least 3 are carp
Pr(exactly 2 are trout) = ?

At least 3 carps have been caught:
ccctttt
cccctttt
cccccttt
cccccctt
ccccccct
cccccccc

Pr(exactly 2 trout|least 3 carps) = 1/6.
The "unpack" suggestion was only if order mattered.

 

[Agent Smith]

@logistic_guy slept through the most crucial stage in my academic life.

 

[nasi112]

If A is a subset of B, then A and B = A. Do you see this?
That's why I could say

Whenever you catch exactly 2 trout, you have caught at most (that is, no more than) 5 trout. So "exactly 2 trout" if a subset of "at most 5 trout".

Here is an example of what I said: If X = {A, B, C} and Y = {A, B, C, 0, 1}, then X and Y = {A, B, C} = X.

I think I did.

Thanks doctor. I think I got your point.

There are only trout and carp in the lake.
8 fish were caught
At least 3 are carp
Pr(exactly 2 are trout) = ?

At least 3 carps have been caught:
ccctttt
cccctttt
cccccttt
cccccctt
ccccccct
cccccccc

Pr(exactly 2 trout|least 3 carps) = 1/6.
The "unpack" suggestion was only if order mattered.

Thanks Agent Smith. I just realized now your answer makes sense.

cccccctt is exactly 2 trouts and it is one of the 6 possibilities so the answer is 1/6.

In this case the unpack suggestion makes sense because order matters.

If order doesn't matters cccccctt is just one of 28 ways the fisherman catches exactly 2 trouts.
cccttttt is just one of 56 ways the fisherman catches 5 trouts.
cccctttt is just one of 70 ways the fisherman catches 4 trouts.
cccccttt is just one of 56 ways the fisherman catches 3 trouts.
cccccctt is just one of 28 ways the fisherman catches 2 trouts.
ccccccct is just one of 8 ways the fisherman catches 1 trout.
cccccccc is the only 1 way the fisherman catches 0 trout.

28/(56 + 70 + 56 + 28 +8 +1) = 0.128 [imath]\neq[/imath] 0.239