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Math Help - Confirm answer: probability question

  1. #1
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    Confirm answer: probability question

    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?
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  2. #2
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    Quote Originally Posted by Runty View Post
    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.
    Last edited by Archie Meade; October 1st 2010 at 10:30 AM.
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  3. #3
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    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?
    Last edited by mr fantastic; September 28th 2010 at 04:19 PM.
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  4. #4
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    Quote Originally Posted by Runty View Post
    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.
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  5. #5
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    Quote Originally Posted by Runty View Post
    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.
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  6. #6
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    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.
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